标签：sicily   高精度乘法   

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Give two positive integers a and b, please help us calculate a*b.

Input

The first line of the input is a positive integer T. T is the number of test cases followed.

Each test case contain two integer a,b (0<=a<=10^100, 0<=b<=10,000) given in one line.

Output

The output of each test case should consist of one line, contain the result of a*b.

Sample Input

12 7

Sample Output

14

Problem Source

Algorithm Course Examination 2006

没什么好说的，大数据乘法，花点时间就能过了。注意输出的问题就好了，比如答案是9，你不能输出000009。 还有输出0，你不能输出0000000。

#include<bits/stdc++.h>
using namespace std;
//string ans;
char tmp[1002][1002];

//大数据加法 
void add(char a[],char b[],char c[]){
	if(strlen(a)<strlen(b))
	swap(a,b);
	int carry=0;
	c[strlen(a)+1]='\0';
	for(int i=strlen(a)-1,j=strlen(b)-1;i>=0;i--,j--){
		int sum=0;
		sum+=(a[i]-'0')+carry;
		if(j>=0)
		sum+=(b[j]-'0');
		
		c[i+1]=sum%10+'0';
		carry=sum/10;
	}
	c[0]=carry+'0';
} 
//乘法 
void mul(char a[],char b[],char c[]){
	if(strlen(a)<strlen(b))
	swap(a,b);
	int len_small = strlen(b);
	int len_big = strlen(a);
	int countt=0;
	
	for(int i=len_small-1;i>=0;i--){
		int carry=0;
		for(int j=len_big-1;j>=0;j--){
			int sum=(a[j]-'0')*(b[i]-'0')+carry;
			tmp[countt][j+1]=sum%10+'0';
			carry=sum/10;
		}
		tmp[countt][0]=carry+'0';
		//补零
		for(int k=1;k<countt+1;k++) tmp[countt][len_big+k]='0'; 
		tmp[countt][len_big+countt+1]='\0';
		countt++;
	}
	char ans[1002]="0";
	char temp[1002];
	for(int i=0;i<countt;i++){
		add(tmp[i],ans,temp);
		strcpy(ans,temp);
	}
	int pos=0;
	int len = strlen(ans); 
	//过滤前导0 
	for(int i=0;i<len;i++)
	if(ans[i]!='0'){
	pos=i;break;
	}
	if(pos==0)
	printf("0");
	else
	for(int i=pos;i<len;i++){
		printf("%c",ans[i]);
		
	}
	printf("\n");
}

int main(){
	int t;
	scanf("%d",&t);
	char a[1002],b[1002],ans[1002];
	while(t--){
		scanf("%s %s",a,b);
		mul(a,b,ans);
	}
}

sicily 1381. a*b

标签：sicily   高精度乘法   

原文地址：http://blog.csdn.net/yuhao199555/article/details/46065427