SDUTOJ 2613 This is an A+B Problem——大数模拟

发布时间：2020-12-14 03:31:25 所属栏目：大数据来源：网络整理

导读：?? This is an A+B Problem Time Limit: 1000ms?? Memory limit: 65536K??有疑问？点这里^_^ 题目描述 As usual,there will be? an A+B ?problem in warming up ,this problem is: Given two integers A and B,your job is to calculate the sum of A + B.

This is an A+B Problem

Time Limit: 1000ms?? Memory limit: 65536K??有疑问？点这里^_^

题目描述

As usual,there will be?an A+B?problem in warming up,this problem is:
Given two integers A and B,your job is to calculate the sum of A + B.

输入

?Th ere are several test cases,For each test case:
There are two integers A,B for each case (0?≤?A,B < 10¹⁰⁰⁰).

输出

?For each test case,output one line containing the result of A+B.?

示例输入

1 2
22222222221 22222222221

示例输出

3
22222222222

提示

大数模拟：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char s1[1100];
char s2[1100];
int ls[1110];

int main()
{
    int l1,l2,i,j,y,a,b;
    memset(s1,'',sizeof(s1));
    memset(s2,sizeof(s2));
    memset(ls,sizeof(ls));
    while(~scanf("%s%s",s1,s2))
    {
        l1=strlen(s1);
        l2=strlen(s2);
        if(l1==1&&s1[0]=='0'&&l2==1&&s2[0]=='0')
        {
            printf("0n");
            continue;
        }                                        //处理0+0=0的情况
        y=0;
        for(i=l1-1,j=l2-1; i>=0||j>=0; i--,j--)
        {
            if(i>=0)
            {
                a=s1[i]-'0';
            }
            else
            {
                a=0;
            }
            if(j>=0)
            {
                b=s2[j]-'0';
            }
            else
            {
                b=0;
            }
            ls[y++]=a+b;
        }
        for(i=0; i<y; i++)
        {
            ls[i+1]+=(ls[i]/10);
            ls[i]%=10;
        }
        i=y;
        while(ls[i]==0)
        {
            i--;
        }
        for(; i>=0; i--)
        {
            printf("%d",ls[i]);
        }
        printf("n");
        memset(s1,sizeof(s1));
        memset(s2,sizeof(s2));
        memset(ls,sizeof(ls));
    }

    return 0;
}

（编辑：李大同）

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