杭电acm 1002,有点崩溃,老是出错,感觉格式什么的都对呀

#include
#include
int main(){
char ax[20][1000], bx[20][1000], temp[1000], a[1000], b[1000];
int n, lena, lenb;
while(scanf("%d",&n)!= EOF){

for(int i = 0;i < n;i++){
    scanf("%s %s",ax[i],bx[i]);
} 
for(int i = 0;i < n;i++){
    int num[1001] = {0};
    strcpy(a,ax[i]);
    strcpy(b,bx[i]);
    lena = strlen(a);
    lenb = strlen(b);
    if(lena < lenb){
        strcpy(temp,a);
        strcpy(a,b);
        strcpy(b,temp);
        lena = strlen(a);
        lenb = strlen(b);
    }
    int carry = 0;
    int len = lena;
    int add = 0;
    for(int j = 0;j < len;j++){
        if(lenb > 0){
            add = a[lena-1] - '0' + b[lenb-1] - '0' + carry;
            num[lena] = add % 10;
            carry = add / 10;
            lena--;
            lenb--;
        }
        else{
            add = a[lena-1] - '0' + carry;
            num[lena] = add % 10;
            carry = add / 10;
            lena--;
            lenb--;
        }   
    }
    num[0] = carry;
    printf("case %d:\n",i+1);
    printf("%s + %s = ",a,b);
    if(num[0] == 0){
        for(int j = 1;j <= len;j++){
            printf("%d",num[j]);
        }
        printf("\n");
    }
    else{
        for(int j = 0;j <= len;j++){
            printf("%d",num[j]);
        }
        printf("\n");
    }
    printf("\n");
}

}
return 0;

}

经典的大数相加,题目如下:

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input
2
1 2
112233445566778899 998877665544332211

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

程序跑图贴在下面,大牛们帮看看吧

4个回答

图片说明

你的出错信息是什么?是wrong还是PE?如果是PE的话,你有可能是输出格式不太对,比如多了个空格或者回车什么的。
如果你是搞ACM的话,我可以给点建议:将答案定向到txt文本里,这样更方便与标准答案对比,避免出现PE这样的低级错误。

 #include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{

    int t,i,x,num=0;
    scanf("%d",&t);
    getchar();
    char a1[1010],b1[1010];
    int a[1010],b[1010],c[1010],lena,lenb,lenc;
    while(scanf("%s%s",&a1,&b1)==2)
    {
        num++;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        lena=strlen(a1);
        lenb=strlen(b1);
        for(i=0;i<=lena;i++)
        a[lena-i]=a1[i]-48;
        for(i=0;i<=lenb;i++)
        b[lenb-i]=b1[i]-48;
        lenc=1,x=0;
        while(lenc<=lena || lenc<=lenb)
        {
            c[lenc]=a[lenc]+b[lenc]+x;
            x=c[lenc]/10;
            c[lenc]%=10;
            lenc++;
        }
        if(c[lenc]==0)
        {
            lenc--;
        }
        printf("Case %d:\n",num);
        printf("%s + %s = ",a1,b1);
        for(i=lenc;i>=1;i--)
        cout<<c[i];
        cout<<endl;
        if(num!=t)
        cout<<endl;
    }
    return 0;
 } 

所以参考我的代码,你的错误在:最后一个测试点你也输出了空格换行,题目说的是两个测试点之间才有空格换行。

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