【bzoj 1754】【POJ - 2389 】Bull Math （高精度运算）
题干：
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.
Input
* Lines 1..2: Each line contains a single decimal number.
Output
* Line 1: The exact product of the two input lines
Sample Input
11111111111111 1111111111
Sample Output
12345679011110987654321
解题报告：
好多错误啊一开始写的时候。就是算个大数乘法。
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX =1000 + 5 ;
char s1[MAX],s2[MAX];
int a[MAX],b[MAX],ans[MAX];
int len1,len2;
int p,c;//p代表当前位置，c代表进位
int main()
{
// freopen("in.txt","r",stdin);
scanf("%s",s1);
scanf("%s",s2);
len1=strlen(s1);len2=strlen(s2);
//
for(int i = 1; i<=len1; i++) {
a[len1-(i-1)]=s1[i-1]-'0';
}
for(int i = 1; i<=len2; i++) {
b[len2-(i-1)]=s2[i-1]-'0';
}
// for(int i = len1; i>=1; i--) printf("%d",a[i]);
// printf("\n");
// for(int i = len2; i>=1; i--) printf("%d",b[i]);
// printf("\n");
for(int i = 1; i<=len2; i++) {
for(int j = 1; j<=len1; j++) {
ans[i+j-1] += b[i]*a[j];
}
}
for(int i = 1; i<=len1+len2; i++) {
p=i;
if(ans[i]<=9) continue;
// c=ans[i]/10;
// while(c>0) {
// ans[++p] += c;
// ans[p-1]=ans[p-1]%10;
// c= ( c+ans[p] )/10;
//
// }
while(ans[i]>9) {
ans[i]-=10;
ans[i+1]++;
}
}
// printf("p = %d\n",p);
for(int i = len1+len2; i>=1; i--) {
if(ans[i]!=0) {
p=i;break;
}
}
for(int i = p; i>=1; i--) {
printf("%d",ans[i]);
}
printf("\n");
return 0 ;
}
//bzoj 1754 [Usaco2005 qua]Bull Math
//11111111111111
//1111111111
