查找元素
找x
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n;int a[200];while (scanf("%d", &n) != EOF){int flag = 0;for (int i = 0; i < n; i++){scanf("%d", &a[i]);}int x;scanf("%d", &x);for (int i = 0; i < n; i++){if (x == a[i]){printf("%d\n", i);flag = 1;}}if (!flag)printf("-1\n");}return 0;}
用的是遍历的方法。
统计同成绩学生人数
题目描述
读入N名学生的成绩，将获得某一给定分数的学生人数输出。
输入
测试输入包含若干测试用例，每个测试用例的格式为
第1行：N
第2行：N名学生的成绩，相邻两数字用一个空格间隔。
第3行：给定分数
当读到N=0时输入结束。其中N不超过1000，成绩分数为（包含）0到100之间的一个整数。
输出
对每个测试用例，将获得给定分数的学生人数输出。
查找学生信息
题目描述
输入N个学生的信息，然后进行查询。
输入
输入的第一行为N，即学生的个数(N<=1000)
接下来的N行包括N个学生的信息，信息格式如下：
01 李江 男 21
02 刘唐 男 23
03 张军 男 19
04 王娜 女 19
然后输入一个M(M<=10000),接下来会有M行，代表M次查询，每行输入一个学号，格式如下：
02
03
01
04
输出
输出M行，每行包括一个对应于查询的学生的信息。
如果没有对应的学生信息，则输出“No Answer!”
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n,m;struct stu{char num[8];char name[100];char sex[20];int age;} student[1000]; //char search[10000][3];char search[8];while (scanf("%d", &n) != EOF){for (int i = 0; i < n; i++){scanf("%s %s %s %d", student[i].num, student[i].name,student[i].sex,&student[i].age);}scanf("%d", &m);for (int i = 0; i < m; i++){scanf("%s", &search);for (int j = 0; j < n; j++){if (strcmp(student[j].num, search)==0){printf("%s %s %s %d\n", student[j].num, student[j].name, student[j].sex, student[j].age);break;}if(j==n-1)printf("No Answer!\n");}}}return 0;}
查找
题目描述
输入数组长度 n
输入数组 a[1…n]
输入查找个数m
输入查找数字b[1…m]
输出 YES or NO 查找有则YES 否则NO 。
输入
输入有多组数据。
每组输入n，然后输入n个整数，再输入m，然后再输入m个整数（1<=m<=n<=100）。
输出
如果在n个数组中输出YES否则输出NO。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n, m;int a[100], b[100];while (scanf("%d", &n) != EOF){for (int i = 0; i < n; i++){scanf("%d", &a[i]);}scanf("%d", &m);for (int j = 0; j < m; j++){scanf("%d", &b[j]);for (int i = 0; i < n; i++){if (a[i] == b[j]){printf("YES\n");break;}if (i == n - 1)printf("NO\n");}}}return 0;}
跟奥巴马一起编程
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n;char c;scanf("%d %c", &n, &c);int m = floor(n / 2);for (int i = 0; i < m; i++){if (i == 0 || i == m - 1){for (int j = 0; j < n; j++){printf("%c", c);if (j == n - 1)printf("\n");}}else{for (int j = 0; j < n; j++){if (j == 0)printf("%c", c);if(j == n - 1)printf("%c\n", c);if(j!=0&&j!=n-1)printf(" ");}}}return 0;}
其中if语句用for循环语句代替可能更好
输出梯形
题目描述
输入一个高度h，输出一个高为h，上底边为h的梯形。
输入
一个整数h(1<=h<=1000)。
输出
h所对应的梯形。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int h;while (scanf("%d", &h) != EOF){int n = 3 * h - 2;for (int i = 0; i < h; i++){for (int j = 0; j < (n-h-2*(i-0)); j++){printf(" ");}for (int j = (n - h - 2 * (i - 0))+1; j <= n; j++){printf("*");if (j == n)printf("\n");}}}return 0;}
Hello World for U
Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, “helloworld” can be printed as:
h d
e l
l r
lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible – that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){char a[80] = {' '};scanf("%s", a);int len;len = strlen(a);int side, mid;side = floor((len + 2) / 3);mid = len - 2 * side;for (int i = 0; i < side-1; i++){printf("%c", a[i]);for (int j = 0; j < mid; j++){printf(" ");}printf("%c\n", a[len - i - 1]);}for (int j = 0; j < mid+2; j++){printf("%c", a[side - 1+j]);}printf("\n");return 0;}
** 等腰梯形**
题目描述
请输入高度h，输入一个高为h，上底边长为h 的等腰梯形（例如h=4，图形如下）。
输入
输入第一行表示样例数m，接下来m行每行一个整数h，h不超过10。
输出
对应于m个case输出要求的等腰梯形。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n;int h;scanf("%d", &n);while (n--){scanf("%d", &h);for (int i = 0; i < h; i++){for (int j = 0; j < h - i - 1; j++){printf(" ");}for (int j = h - i - 1; j < 2*h+i-1; j++){printf("*");}for (int j = 2*h+i-1; j < 3*h-2; j++){printf(" ");}printf("\n");}}return 0;}
沙漏图形 tri2str [1 * +]
问题：输入n，输出正倒n层星号三角形。首行顶格，星号间有一空格，效果见样例 输入样例： 3 输出样例：* * ** * ** * * * *数据规模 1<= n <=50
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int n;int h;scanf("%d", &n);for (int i = 0; i < n; i++){for (int j = 0; j < i; j++){printf(" ");}for (int j = 0; j < n-i-1; j++){printf("* ");}printf("*\n");}for (int i = 0; i < n-1; i++){for (int j = 0; j < n - 2 - i; j++){printf(" ");}for (int j = 0; j < i+1; j++){printf("* ");}printf("*\n");}return 0;}
日期处理
日期差值
// ConsoleApplication1.cpp: 定义控制台-应用程序的入口点。//#include "stdafx.h"#include <cstdio>#include <cstring>#include <cmath>using namespace std;void change(int* p, int* q){if (*p >= *q){int temp;temp = *p;*p = *q;*q = temp;}}int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int time1, time2, sum = 1;int tempyear;while (scanf("%d%d", &time1, &time2) != EOF){int *p = &time1, *q = &time2;change(p, q);int year1, year2, month1, month2, day1, day2;year1 = floor(time1 / 10000);year2 = floor(time2 / 10000);month1 = time1 % 10000 / 100;month2 = time2 % 10000 / 100;day1 = time1 % 100;day2 = time2 % 100;//int monthday[2][13] =//{ 0,0,31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31 };//上一行为平年，下一行为闰年int monthday[2][13] ={ 0,31,27,31,30,31,30,31,31,30,31,30,31,0,31,28,31,30,31,30,31,31,30,31,31,30 };while (year1 < year2-1){if (Isleap(year1) == 1){sum += 366;year1++;}else{year1++;sum += 365;}}while (month1 <= 12){sum += monthday[Isleap(year1)][month1];month1++;if (month1 == 13){month1 = 1;year1++;break;}}while (year1<year2||month1<month2||day1 < day2){sum++;day1++;if (day1 == monthday[Isleap(year1)][month1]){day1 = 1;month1++;}if (month1 == 13){month1 = 1;year1++;}}printf("%d", sum);}return 0;}
！！！！注意！以上代码至少是时间超限的（还有可能错误）。累了==
讨论版有个方法是吧所有的单位设置为日期，是个好想法
试试看：
#include <cstdio>#include <cstring>#include <cmath>using namespace std;void change(int* p, int* q){if (*p >= *q){int temp;temp = *p;*p = *q;*q = temp;}}int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int time1, time2, sum = 1;int tempyear;while (scanf("%d%d", &time1, &time2) != EOF){int *p = &time1, *q = &time2;change(p, q);int year1, year2, month1, month2, day1, day2;year1 = floor(time1 / 10000);year2 = floor(time2 / 10000);month1 = time1 % 10000 / 100;month2 = time2 % 10000 / 100;day1 = time1 % 100;day2 = time2 % 100;//int monthday[2][13] =//{ 0,0,31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31 };//上一行为平年，下一行为闰年int monthday[2][13] ={ 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,31,30 };int turn1 = 0, turn2 = 0;for (int month = 1; month < month1; month++){turn1 += monthday[Isleap(year1)][month];}int k = 1;for (int day = 1; day <= day1; day++){turn1++;}k = 2;for (int year = year1; year < year2; year++){if (Isleap(year) == 1)turn2 += 366;elseturn2 += 365;}k = 3;for (int month = 1; month < month2; month++){turn2 += monthday[Isleap(year2)][month];}k = 4;for (int day = 1; day <= day2; day++){turn2++;}k = 5;sum = turn2 - turn1 + 1;printf("%d\n", sum);}return 0;}
我踏马终于做对了
Day of Week
We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.
For example, years , 2180 and 2400 are leap. Years , 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.
输入
There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.
输出
Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int time1, turn=0;int year1, month1, day1;char monthc[15];while (scanf("%d %s %d", &day1,monthc,&year1) != EOF){turn = 0;if (strcmp(monthc, "January") == 0)month1 = 1;if (strcmp(monthc, "February") == 0)month1 = 2;if (strcmp(monthc, "March") == 0)month1 = 3;if (strcmp(monthc, "April") == 0)month1 = 4;if (strcmp(monthc, "May") == 0)month1 = 5;if (strcmp(monthc, "June") == 0)month1 = 6;if (strcmp(monthc, "July") == 0)month1 = 7;if (strcmp(monthc, "August") == 0)month1 = 8;if (strcmp(monthc, "September") == 0)month1 = 9;if (strcmp(monthc, "October") == 0)month1 = 10;if (strcmp(monthc, "November") == 0)month1 = 11;if (strcmp(monthc, "December") == 0)month1 = 12;int monthday[2][13] ={ 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };for (int year = 1000; year < year1; year++){if (Isleap(year) == 1)turn += 366;elseturn += 365;}for (int month = 1; month < month1; month++){turn += monthday[Isleap(year1)][month];}turn += day1;turn = (turn-1) % 7;switch (turn){case 0:{printf("Wednesday\n");break;}case 1:{printf("Thursday\n");break;}case 2:{printf("Friday\n");break;}case 3:{printf("Saturday\n");break;}case 4:{printf("Sunday\n");break;}case 5:{printf("Monday\n");break;}case 6:{printf("Tuesday\n");break;}}}return 0;}
拓展：有一个公式叫基姆拉尔森计算公式，可以通过日期判断是星期几。
打印日期
题目描述
给出年分m和一年中的第n天，算出第n天是几月几号。
输入
输入包括两个整数y(1<=y<=3000)，n(1<=n<=366)。
输出
可能有多组测试数据，对于每组数据，按 yyyy-mm-dd的格式将输入中对应的日期打印出来。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int time1, turn ,sum;int year1, month1, day1;while (scanf("%d %d", &year1, &turn) != EOF){sum = 0;int monthday[2][13] ={ 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };for (int month = 1; month <= 12; month++){sum += monthday[Isleap(year1)][month];int k = 1;if (sum > turn){sum -= monthday[Isleap(year1)][month];month1 = month;day1 = turn - sum;printf("%04d-%02d-%02d\n", year1, month1, day1);break;}if (sum == turn){day1 = monthday[Isleap(year1)][month];month1 = month--;printf("%04d-%02d-%02d\n", year1, month1, day1);break;}}}return 0;}
日期类
题目描述
编写一个日期类，要求按xxxx-xx-xx 的格式输出日期，实现加一天的操作。
输入
输入第一行表示测试用例的个数m，接下来m行每行有3个用空格隔开的整数，分别表示年月日。测试数据不会有闰年。
输出
输出m行。按xxxx-xx-xx的格式输出，表示输入日期的后一天的日期。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int year1, month1, day1;int monthday[2][13] ={ 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };int m;while (scanf("%d", &m) != EOF){while (m--){scanf("%d %d %d", &year1, &month1, &day1);day1++;if (day1 == monthday[Isleap(year1)][month1]+1){day1 = 1;month1++;}if (month1 == 12){month1 = 1;year1++;}printf("%04d-%02d-%02d\n", year1, month1, day1);}}return 0;}
日期累加
题目描述
设计一个程序能计算一个日期加上若干天后是什么日期。
输入
输入第一行表示样例个数m，接下来m行每行四个整数分别表示年月日和累加的天数。
输出
输出m行，每行按yyyy-mm-dd的个数输出。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int Isleap(int year){if ((year % 400 == 0)|| (year % 4 == 0 && year % 100 != 0)){return 1;}elsereturn 0;}int main(){int year1, month1, day1,turn;int monthday[2][13] ={ 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };int m;while (scanf("%d", &m) != EOF){while (m--){scanf("%d %d %d %d", &year1, &month1, &day1,&turn);while(turn--){day1++;if (day1 == monthday[Isleap(year1)][month1] + 1){day1 = 1;month1++;}if (month1 == 12+1){month1 = 1;year1++;}if (turn == 0){printf("%04d-%02d-%02d\n", year1, month1, day1);break;}}}}return 0;}
进制转换
两种不同进制进行转换
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int mtop(int m , int a){int p = 0,product=1;while (a != 0){p += (a % 10)*product;a /= 10;product *= m;int k = 1;}return p;}int pton(int n, int temp, int *p0){int i = 0;do {*p0 =temp % n;temp /= n;p0++;i++;} while (temp != 0);return i;}int main(){int a,temp,m,n,len;int b[50] = { 0 };int* p0=b;scanf("%d%d", &m, &n);scanf("%d", &a);temp=mtop(m,a);printf("%d\n", temp);len=pton(n,temp,p0);for (int i = len-1; i >= 0; i--){printf("%d", b[i]);}printf("\n");return 0;}
D进制的A+B
输入非负十进制整数AB以及D（进制数），输出A+B的D进制结果
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int mtop(int m, int a){int p = 0, product = 1;while (a != 0){p += (a % 10)*product;a /= 10;product *= m;int k = 1;}return p;}int pton(int n, int temp, int *p0){int i = 0;do{*p0 = temp % n;temp /= n;p0++;i++;} while (temp != 0);return i;}int main(){int num1, num2, N,len;scanf("%d%d%d", &num1, &num2, &N);int b[50] = { 0 };int* p0 = b;int sum, turnsum;sum = num1 + num2;len = pton(N, sum, p0);for (int i = len - 1; i >= 0; i--){printf("%d", b[i]);}printf("\n");return 0;}
又一版 A+B
题目描述
输入两个不超过整型定义的非负10进制整数A和B(<=231-1)，输出A+B的m (1 < m <10)进制数。
输入
输入格式：测试输入包含若干测试用例。每个测试用例占一行，给出m和A，B的值。
当m为0时输入结束。
输出
输出格式：每个测试用例的输出占一行，输出A+B的m进制数。
提示
注意输入的两个数相加后的结果可能会超过int和long的范围。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int mtop(int m, int a){int p = 0, product = 1;while (a != 0){p += (a % 10)*product;a /= 10;product *= m;int k = 1;}return p;}int pton(int n, long long temp, int *p0){int i = 0;do{*p0 = temp % n;temp /= n;p0++;i++;} while (temp != 0);return i;}int main(){int N;while (scanf("%d", &N),N != 0){long long num1, num2, len;scanf("%lld%lld",&num1, &num2 );int b[5000] = { 0 };int* p0 = b;long long sum, turnsum;sum = num1 + num2;len = pton(N, sum, p0);for (int i = len - 1; i >= 0; i--){printf("%d", b[i]);}printf("\n");}return 0;}
** 数制转换**
题目描述
求任意两个不同进制非负整数的转换（2进制～16进制），所给整数在long所能表达的范围之内。
不同进制的表示符号为（0，1，…，9，a，b，…，f）或者（0，1，…，9，A，B，…，F）。
输入
输入只有一行，包含三个整数a，n，b。a表示其后的n 是a进制整数，b表示欲将a进制整数n转换成b进制整数。a，b是十进制整数，2 =< a，b <= 16。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int a, b;char str[500];long long n;while (scanf("%d%s%d", &a, str, &b) != EOF){//转换成十进制// 转换成int类型int lenth_of_a = strlen(str);int p = 0;int product = 1;for (int i = lenth_of_a-1; i >= 0; i--){switch (str[i]){case '0': {p += 0 * product; break; }case '1': {p += 1 * product; break; }case '2': {p += 2 * product; break; }case '3': {p += 3 * product; break; }case '4': {p += 4 * product; break; }case '5': {p += 5 * product; break; }case '6': {p += 6 * product; break; }case '7': {p += 7 * product; break; }case '8': {p += 8 * product; break; }case '9': {p += 9 * product; break; }case 'a': {p += 10 * product; break; }case 'A': {p += 10 * product; break; }case 'b': {p += 11 * product; break; }case 'B': {p += 11 * product; break; }case 'c': {p += 12 * product; break; }case 'C': {p += 12 * product; break; }case 'd': {p += 13 * product; break; }case 'D': {p += 13 * product; break; }case 'e': {p += 14 * product; break; }case 'E': {p += 14 * product; break; }case 'f': {p += 15 * product; break; }case 'F': {p += 15 * product; break; }default:break;}product *= a;int rk = 0;}//转换成b进制char output[500];int i = 0;int temp;do {temp = p % b;switch (temp){case 0: {output[i] = '0';break; }case 1: {output[i] = '1'; break; }case 2: {output[i] = '2'; break; }case 3: {output[i] = '3';break; }case 4: {output[i] = '4'; break; }case 5: {output[i] = '5'; break; }case 6: {output[i] = '6'; break; }case 7: {output[i] = '7'; break; }case 8: {output[i] = '8'; break; }case 9: {output[i] = '9'; break; }case 10: {output[i] = 'A'; break; }case 11: {output[i] = 'B'; break; }case 12: {output[i] = 'C'; break; }case 13: {output[i] = 'D'; break; }case 14: {output[i] = 'E'; break; }case 15: {output[i] = 'F'; break; }default:break;}p/= b;i++;} while (p != 0);for (int j = i - 1; j >= 0; j--){printf("%c", output[j]);}printf("\n");}return 0;}
进制转换
遇到问题题目描述
将一个长度最多为30位数字的十进制非负整数转换为二进制数输出。
输入
多组数据，每行为一个长度不超过30位的十进制非负整数。
（注意是10进制数字的个数可能有30个，而非30bits的整数）
输出
每行输出对应的二进制数。
？？？？？？？？？？？？？？？？
八进制
题目描述
输入一个整数，将其转换成八进制数输出。
输入
输入包括一个整数N(0<=N<=100000)。
输出
可能有多组测试数据，对于每组数据，
输出N的八进制表示数。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){long long p;while (scanf("%lld", &p) != EOF){char output[500];int i = 0;int temp;do {temp = p % 8;switch (temp){case 0: {output[i] = '0'; break; }case 1: {output[i] = '1'; break; }case 2: {output[i] = '2'; break; }case 3: {output[i] = '3'; break; }case 4: {output[i] = '4'; break; }case 5: {output[i] = '5'; break; }case 6: {output[i] = '6'; break; }case 7: {output[i] = '7'; break; }case 8: {output[i] = '8'; break; }default:break;}p /= 8;i++;} while (p != 0);for (int j = i - 1; j >= 0; j--){printf("%c", output[j]);}printf("\n");}return 0;}
【字符串】回文串
题目描述
读入一串字符，判断是否是回文串。“回文串”是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。
输入
一行字符串，长度不超过255。
输出
如果是回文串，输出“YES”，否则输出“NO”。
#include <cstdio>#include <cstring>#include <cmath>const int maxn = 256;using namespace std;bool judge(char str[]){int len = strlen(str);for (int i = 0; i < len / 2; i++){if (str[i] != str[len - i - 1])return false;}return true;}int main(){char str[maxn];//while (gets(str))scanf("%s", str);bool flag = judge(str);if (flag == true)printf("YES\n");elseprintf("NO\n");return 0;}
单词倒序输出
#include <cstdio>#include <cstring>#include <cmath>const int maxn = 3000;using namespace std;int main(){int num = 0;char ans[90][90];/*while (scanf("%s", ans[num] )!= EOF){num++;}*/char str[90];fgets(str,90,stdin);int len = strlen(str), r = 0, h = 0;for (int i = 0; i <= len; i++){if(str[i]!='\n'){if (str[i] != ' '){ans[r][h++] = str[i];}else{ans[r][h] = '\0';r++;h = 0;}}}for (int i = r; i >=0 ; i--){printf("%s", ans[i]);if (i > 0)printf(" ");}return 0;}
注意题目中的’\n’和’\0’
字符串连接
题目描述
不借用任何字符串库函数实现无冗余地接受两个字符串，然后把它们无冗余的连接起来。
输入
每一行包括两个字符串，长度不超过100。
输出
可能有多组测试数据，对于每组数据，
不借用任何字符串库函数实现无冗余地接受两个字符串，然后把它们无冗余的连接起来。
输出连接后的字符串。
#include <cstdio>#include <cstring>#include <cmath>//const int maxn = 200;using namespace std;int main(){char str1[200], str2[100];char str3[100];int len, len1, len2;while (scanf("%s%s", str1, str2) != EOF){len1 = strlen(str1);len2 = strlen(str2);for (int i = len1; i < len1+len2; i++){str1[i] = str2[i - len1];int k = 0;}str1[len1 + len2] = '\0';printf("%s\n", str1);}}
首字母大写
题目描述
对一个字符串中的所有单词，如果单词的首字母不是大写字母，则把单词的首字母变成大写字母。
在字符串中，单词之间通过空白符分隔，空白符包括：空格(’ ‘)、制表符(’\t’)、回车符(’\r’)、换行符(’\n’)。
输入
输入一行：待处理的字符串（长度小于100）。
输出
可能有多组测试数据，对于每组数据，
输出一行：转换后的字符串。
#include <cstdio>#include <cstring>#include <cmath>//const int maxn = 200;using namespace std;int main(){char str[105];int len;while (fgets(str,110,stdin)!=NULL){len = strlen(str);for (int i = 0; i < len; i++){if (i == 0&&str[i]>=97&&str[i]<=122){str[i] -= 32;//要不要加单引号？}if (str[i] == ' ' || str[i] == '\t' || str[i] == '\r' || str[i] == '\n'){if (str[i + 1] >=97 && str[i + 1] <= 122)str[i + 1] = str[i + 1] - 32;}}printf("%s", str);}return 0;}
字符串的查找删除
答案错误？？？？
#include <cstdio>#include <cstring>#include <cmath>//const int maxn = 200;using namespace std;int judge1(char str[], char text[], int i, int len1){for (int j = 0; j < len1; j++){char qq= str[j];char qqr = text[i];if (str[j] == text[i] || (str[j] <= 122 && str[j] >= 97 && ((str[j] - 32) == text[i])) || (str[j] <= 90 && str[j] >= 65 && ((str[j] + 32) == text[i]))){int nothing=5;i++;}elsereturn 0;}int ee = 5;return 1;}int main(){char str[500];char text[5000];scanf("%s", str);getchar();while (fgets(text, 5000, stdin) != NULL){int len0 = strlen(text);int len1 = strlen(str);for (int i = 0; i < len0 - len1 + 1; i++){int flag = judge1(str, text, i, len1);if (flag){i = i + len1-1;int k = 0;}if (flag==0&&text[i] != ' '){printf("%c", text[i]);int m = 1;}}printf("\n");}return 0;}
** 单词替换**
不知道错在哪题目描述
输入一个字符串，以回车结束（字符串长度<=100）。该字符串由若干个单词组成，单词之间用一个空格隔开，所有单词区分大小写。现需要将其中的某个单词替换成另一个单词，并输出替换之后的字符串。
输入
多组数据。每组数据输入包括3行，
第1行是包含多个单词的字符串 s，
第2行是待替换的单词a，(长度<=100)
第3行是a将被替换的单词b。(长度<=100)
s, a, b 最前面和最后面都没有空格。
输出
每个测试数据输出只有 1 行，
将s中所有单词a替换成b之后的字符串。
#include <cstdio>#include <cstring>#include <cmath>//const int maxn = 200;using namespace std;int main(){char str[500];char output[500][500];char a[110], b[110];while (fgets(str,500,stdin)!=NULL){scanf("%s\n%s", a,b);//要不要\n?int len = strlen(str);int r = 0, h = 0;for (int i = 0; i < len; i++){if (str[i] == ' '||str[i]=='\n'){output[r][h] = '\0';r++;h = 0;}else{output[r][h] = str[i];h++;}}for (int j = 0; j < r; j++){if (!strcmp(output[j], a)){printf("%s ", b);}else{printf("%s", output[j]);if (j == r - 1)printf("\n");else printf(" ");}}}return 0;}
字符串去特定字符
题目描述
输入字符串s和字符c，要求去掉s中所有的c字符，并输出结果。
输入
测试数据有多组，每组输入字符串s和字符c。
输出
对于每组输入,输出去除c字符后的结果。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){char str[100];char c;while (fgets(str,100,stdin)!=NULL){c = getchar();int len = strlen(str);for (int i = 0; i < len; i++){if (str[i] != c)printf("%c",str[i]);}getchar();//printf("\n");}return 0;}
这里一定要注意用getchar()来吸收换行符！
输入可能包含空格，为了使字符串不被中断，函数不可以选用scanf，而应该选用fgets(str,buf,stdin)；由于fgets()函数会把\n也写入字符数组中，所以在这个程序中，打印完字符串后会跟上\n（居然在\0后面？？？），在下一次循环时，直接将\n误读入，结束fgets字符串的输入，导致程序结果错误。
数组逆置
题目描述
输入一个字符串，长度小于等于200，然后将数组逆置输出。
输入
测试数据有多组，每组输入一个字符串。
输出
对于每组输入,请输出逆置后的结果。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){char str[210];int len;while (fgets(str,210,stdin)!=NULL){len = strlen(str);int i;for (i = len-1; i >= 0; --i){putchar(str[i]);}printf("\n");}return 0;}
** 比较字符串**
题目描述
输入两个字符串，比较两字符串的长度大小关系。
输入
输入第一行表示测试用例的个数m，接下来m行每行两个字符串A和B，字符串长度不超过50。
输出
输出m行。若两字符串长度相等则输出A is equal long to B；若A比B长，则输出A is longer than B；否则输出A is shorter than B。
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int m;char a[55], b[55];int lena, lenb;while (scanf("%d", &m) != EOF){while (m--){scanf("%s%s", a, b);lena = strlen(a);lenb = strlen(b);if (lena == lenb)printf("%s is equal long to %s\n", a, b);if (lena > lenb)printf("%s is longer than %s\n", a, b);if (lena < lenb)printf("%s is shorter than %s\n", a, b);}}return 0;}
编排字符串
输入
第一行为字符串个数m，接下来m行每行一个字符床，m不超过100，每个字符床长度不超过20。
输出
输出m行，每行按照样例格式输出，注意用一个空格隔开。
样例输入
5
EricZ
David
Peter
Alan
Jane
样例输出
1=EricZ
1=David 2=EricZ
1=Peter 2=David 3=EricZ
1=Alan 2=Peter 3=David 4=EricZ
1=Jane 2=Alan 3=Peter 4=David
#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int m;char num[110][25];while (scanf("%d", &m) != EOF){for (int i = 0; i < m; i++){scanf("%s", num[i]);for (int j = i; j >= 0; j--){if (j < i - 3)break;printf("%d=%s", i - j + 1, num[j]);if (j == 0||j==i-3)printf("\n");else{printf(" ");}}}}return 0;}
