poj 1504 我想问下为什么会TLE超时？

[cdddfq]

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
int main()
{
    int n,i,j;
    char str[100];
    int fin[104];
    scanf("%d",&n);
    while(n--)
    {
        memset(fin,0,sizeof(fin));
        int len,t=2;
        while(t--){
        scanf("%s",str);
        getchar();
        len=strlen(str);
        int temp=0,c;
        for(i=len-1,j=0;i>=0;i--,j++)
        {
            c=str[i]-'0'+fin[j]+temp;
            temp=c/10;
            fin[j]=c%10;
        }//for
        while(temp>0)
        {
            c=temp+fin[j];
            fin[j]=c%10;
            temp=c/10;
            j++;
        }
        }//while(t)
        int left=0,right=103;
        while(1)
        {
            if(fin[right]==0)
                right--;
            else
                break;
        }
        while(1)
        {
            if(fin[left]==0)
                left++;
            else
                break;
        }
        for(i=left;i<=right;i++)
        {
            printf("%d",fin[i]);
        }
    }
    return 0;
}

[mewiteor]

这题数不大，直接读入字符串反转，转int直接加，再反转清前导0即可

[fly_dragon_fly]

看起来上面模拟加法的地方已经错了，假设了两个数位数一致

[赵4老师]

仅供参考：

#include <stdio.h>
#include <string.h>
#define MAXLEN 1000
char a1[MAXLEN];
char a2[MAXLEN];
static int v1[MAXLEN];
static int v2[MAXLEN];
static int v3[MAXLEN];
int i,j,n,L,z;
void main(void) {
    scanf("%d",&n);
    for (j=0;j<n;j++) {
        scanf("%s%s",a1,a2);

        L=strlen(a1);
        for (i=0;i<L;i++) v1[i]=a1[L-1-i]-'0';

        L=strlen(a2);
        for (i=0;i<L;i++) v2[i]=a2[L-1-i]-'0';

        for (i=0;i<MAXLEN;i++) v3[i]=v1[i]+v2[i];

        for (i=0;i<MAXLEN;i++) {
            if (v3[i]>=10) {
                v3[i+1]+=v3[i]/10;
                v3[i]=v3[i]%10;
            }
        }

        printf("Case %d:\n", j+1);
        printf("%s + %s = ", a1, a2);

        z=0;
        for (i=MAXLEN-1;i>=0;i--) {
            if (z==0) {
                if (v3[i]!=0) {
                    printf("%d",v3[i]);
                    z=1;
                }
            } else {
                printf("%d",v3[i]);
            }
        }
        if (z==0) printf("0");

        printf("\n");
    }
}
//Sample Input
//3
//0 0
//1 2
//112233445566778899 998877665544332211
//
//Sample Output
//Case 1:
//0 + 0 = 0
//Case 2:
//1 + 2 = 3
//Case 3:
//112233445566778899 + 998877665544332211 = 1111111111111111110