BZOJ 1754: [Usaco2005 qua]Bull Math

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros). FJ asks that you do this yourself; don't use a special library function for the multiplication. 输入两个数,输出其乘积

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

题解：

高精度乘法即可。

代码：

#include<cstdio>

#include<cstring>

#include<algorithm>

//by zrt

//problem:

using namespace std;

char a[105],b[105];

int c[105];

int la,lb;

int cc;

int stk[105],top;

int main(){

    #ifdef LOCAL

    freopen("in.txt","r",stdin);

    freopen("out.txt","w",stdout);

    #endif

    scanf("%s%s",a,b);

    la=strlen(a);

    lb=strlen(b);

    for(int i=0;i<la;i++) a[i]-='0';

    for(int i=0;i<lb;i++) b[i]-='0';

    for(int i=0;i<la/2;i++) swap(a[i],a[la-i-1]);

    for(int i=0;i<lb/2;i++) swap(b[i],b[lb-i-1]);

    for(int i=0;i<la;i++){

        for(int j=0;j<lb;j++){

            c[i+j]+=a[i]*b[j];

        }

    }

    for(int i=0;i<=100;i++){

        stk[top++]=(c[i]+cc)%10;

        cc=(c[i]+cc)/10;

    }

    while(stk[top-1]==0) top--;

    while(top){

        printf("%d",stk[--top]);

    }

    puts("");

    return 0;

}

巴特西

BZOJ 1754: [Usaco2005 qua]Bull Math

Description

Input

Output

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