标签：class mit %s 技术 ase exce strstr why should
Input
each test case contains two numbers A and B.
Output
for each case, if A is equal to B, you should print "YES", or print "NO".
Sample Input
1 2 2 2 3 3 4 3
Sample Output
NO YES YES NO
// too young
1 #include<stdio.h> 2 int main() 3 { 4 int a, b; 5 while(scanf("%d %d", &a, &b)!=EOF) 6 { 7 if(a==b) printf("YES\n"); 8 else printf("NO\n"); 9 } 10 return 0; 11 }
// still don‘t know why
1 #include<stdio.h> 2 #include<stdlib.h> 3 int main() 4 { 5 double a,b; 6 char num1[1000000], num2[1000000]; 7 while(scanf("%s %s", num1, num2)!=EOF) 8 { 9 a=atof(num1); b=atof(num2); 10 if(a==b) printf("YES\n"); 11 else printf("NO\n"); 12 } 13 return 0; 14 }
// 小数点后无意义的零要删去
1 #include<stdio.h> 2 #include<string.h> 3 char a[1000000], b[1000000]; 4 void process(char *num) 5 { 6 int len=strlen(num); 7 if(strstr(num,".")!=NULL) 8 { 9 while(num[--len]==‘0‘); 10 if(num[len]==‘.‘) 11 num[len]=‘\0‘; 12 } 13 } 14 int main() 15 { 16 while(scanf(" %s %s", a, b)!=EOF) 17 { 18 process(a); process(b); 19 if(strcmp(a,b)) printf("NO\n"); 20 else printf("YES\n"); 21 } 22 return 0; 23 }
//
1 #include<stdio.h> 2 #include<string.h> 3 char a[1000000], b[1000000]; 4 void process(char *num) 5 { 6 int len=strlen(num); 7 if(strstr(num,".")!=NULL) 8 { 9 while(num[--len]==‘0‘); 10 if(num[len]==‘.‘) 11 len--; 12 num[++len]=‘\0‘; 13 } 14 } 15 int main() 16 { 17 while(scanf(" %s %s", a, b)!=EOF) 18 { 19 process(a); process(b); 20 if(strcmp(a,b)) printf("NO\n"); 21 else printf("YES\n"); 22 } 23 return 0; 24 }
// 判断两数是否相等还可用strcmp
// char *strstr(char *str1, char *str2); 寻找str2在str1中第一次出现的位置，返回位置指针
标签：class mit %s 技术 ase exce strstr why should
原文地址：https://www.cnblogs.com/goldenretriever/p/10355777.html
