acm竞赛例题

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 244704    Accepted Submission(s): 47183

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

Author

Ignatius.L

解题

#include<stdio.h>
#include<string.h>
int main()
{
    int n,m,len1,len2,maxlen,minlen;
    char a[1010],b[1010],sum[1010];
    char *max, *min;
    scanf("%d",&n);
    for(m=1;m<=n;m++)
    {
        int i,j,flag=0;
        memset(sum,0,sizeof(sum));
        scanf("%s %s",a,b);
        len1=strlen(a);
        len2=strlen(b);
        if(len1>=len2)//判断字符串哪个长
        {
            maxlen=len1;
            minlen=len2;
            max = a;
            min = b;
        }
        else
        {
            maxlen=len2;
            minlen=len1;
            max = b;
            min = a;
        }
        for(i=maxlen-1,j=minlen-1;i>=maxlen-minlen&&j>=0;i--,j--)//做循环把两字符串相加放到sum,用flag记录进位,大于9时候为1，并且sum-10，
        {                                                       //只加完到最短的(从后面开始加，方便输出从前面0下表开始输出)
            sum[i]+=max[i]+min[j]-‘0‘-‘0‘;
            if(i==0&&sum[0]>9)
            {
                flag=1;
                sum[0]-=10;
            }
            else if(sum[i]>9)
            {
                sum[i-1]+=1;//进位
                sum[i]-=10;//减10
            }
        }
        for(i=maxlen-minlen-1;i>=0;i--)//再加比最短的长出的部分的,从后面开始加。
        {
            sum[i]+=max[i]-‘0‘;
            if(i==0&&sum[0]>9)
            {
                flag=1;
                sum[i]-=10;
            }
            else if(sum[i]>9)
            {
                sum[i-1]++;//进位
                sum[i]-=10;//减10
            }
        }
        // printf("       %d\n",flag);
        if(flag)                    //输出格式
        {
            printf("Case %d:\n",m);
            printf("%s + %s = 1",a,b);}//a，b字符串不能修改，flag=1，sum[0]=0的时候的进位
        else
        {
            printf("Case %d:\n",m);
            printf("%s + %s = ",a,b);
        }
        for(i=0;i<maxlen;i++)            //循环顺序输出结果
            printf("%d",sum[i]);
        printf("\n");
    }
    return 0;
}

acm竞赛例题

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原文地址：http://my.oschina.net/qiyong/blog/393433