題干：

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

解題報告：

? ? ? 好多錯誤啊一開始寫的時候。就是算個大數乘法。

AC代碼：

#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代表當前位置，c代表進位 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

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