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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15255 Accepted Submission(s): 3917
1 #include <stdio.h> 2 #include <string.h> 3 char str[110] ; 4 int num[100], ans[10010] ; 5 int main() 6 { 7 int i, n, j ; 8 int len, flag ; 9 scanf("%d", &n) ; 10 while(n--) 11 { 12 int total = 0 ; 13 memset(ans, 0, sizeof(ans)) ; 14 while(~scanf("%s", str)) 15 { 16 if(strcmp(str, "0") == 0) 17 break ; 18 total++ ; 19 20 int len = strlen(str) ; 21 for(j=0, i=len-1; i>=0; i--) 22 num[j++] = str[i] - ‘0‘ ; 23 for(i = 0; i < len; i++) 24 ans[i] += num[i] ; 25 for(i=0; i<10009; i++) 26 { 27 if(ans[i] > 9) 28 { 29 ans[i] -= 10 ; 30 ans[i+1] += 1 ; 31 } 32 } 33 } 34 if(total == 0) //只输入一个零 ;WA了很多次; 35 { 36 printf("0\n") ; 37 if(n != 0) 38 printf("\n") ; 39 continue ; 40 } 41 for(i = 10008; i >= 0; i--) 42 { 43 if(ans[i]) 44 break ; 45 } 46 for(; i>=0; i--) 47 printf("%d",ans[i]) ; 48 printf("\n") ; 49 if(n != 0) 50 printf("\n") ; 51 } 52 return 0 ; 53 }
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原文地址:http://www.cnblogs.com/fengshun/p/4674621.html