題干：

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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

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解題報告：

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? ? 水題模擬。

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= ans[p]/10;}//或者用下面的代碼： // 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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錯誤代碼：(有空看看哪里有問題）
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Bull Math

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#include<bits/stdc++.h>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++) {p=i;c=0;for(int j = 1; j<=len1; j++) {ans[p]+=b[i]*a[j];c=ans[p]/10;ans[p]%=10;while(c>0) {ans[++p] += c%10;//這里是+=！！ c/=10;}} for(int i = p; i>=1; i--) {printf("%d",ans[i]);}printf("\n");}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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總結：

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? ? 你在搞笑嗎這么簡單的模擬你搞了一個小時。。。。邏輯關系能不能理清楚了再敲啊。。。

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// printf("p = %d\n",p);

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總結

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