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UVA 10106 Product 高精度运算

时间:2015-08-15 18:25:47      阅读:235      评论:0      收藏:0      [点我收藏+]

 

J - Product
Crawling in process... Crawling failed Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description

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 Product 

The Problem

The problem is to multiply two integers X, Y. (0<=X,Y<10250)

The Input

The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.

The Output

For each input pair of lines the output line should consist one integer the product.

Sample Input

12
12
2
222222222222222222222222

Sample Output

144
444444444444444444444444

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ACcode:

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #define maxn 10000
    using namespace std;
    char a[maxn],b[maxn];
    int ans[maxn*2];
    int main(){
        int lena,lenb;
        while(scanf("%s",a)!=EOF){
            scanf("%s",b);
            if(!strcmp(a,"0")||!strcmp(b,"0")){
                printf("0\n");
                continue;
            }
            memset(ans,0,sizeof(ans));
            lena=strlen(a);
            lenb=strlen(b);
            ans[0]=lena+lenb;
            int t1,t2;
            for(int i=lena;i>0;--i){
                for(int j=lenb;j>0;--j){
                    t1=a[i-1]-'0';
                    t2=b[j-1]-'0';
                    ans[i+j]+=t1*t2;
                }
            }
            for(int i=ans[0];i>0;--i){
                if(ans[i]>10){
                    ans[i-1]+=ans[i]/10;
                    ans[i]%=10;
                }
            }
            int t=1;
            while(ans[t]==0)t++;
            for(int i=t;i<=ans[0];++i)cout<<ans[i];
            putchar('\n');
        }
        return 0;
    }


 

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UVA 10106 Product 高精度运算

原文:http://blog.csdn.net/zp___waj/article/details/47683099

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