Prime Cryptarithm
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http://train.usaco.org/usacoprob2?a=dofOQ4vbPD7&S=crypt1
题不难,题意很重!很重要!
题目大意:
* * * x * * ------- * * * <-- partial product 1 * * * <-- partial product 2 ------- * * * *把给定的数字带入上面乘法竖式后能使竖式成立的式子称为“牛式”
输入一组数字,找出所有的牛式
注意:
partial product 1,partial product 2 都是三位数
#include <iostream>#include <fstream>#include <algorithm>#include <cstring>using namespace std;bool f[10];//可用的数字 int a[10];bool isok(int temp){while(temp){if(!f[temp%10]) return false;temp /= 10;}return true;}int main(){ifstream fin("crypt1.in");ofstream fout("crypt1.out");int n, temp, k, c1, c2, ans;while(fin >> n){memset(f, 0, sizeof(f));ans = k = 0;for(int i = 0; i < n; i++) {fin >> temp;if(!f[temp]){f[temp] = true;a[k++] = temp;}}for(int i = 0; i < k; i++){for(int j = 0; j < k; j++){for(int m = 0; m < k; m++){c1 = a[i]*100 + a[j]*10 + a[m];for(int x = 0; x < k; x++){if(a[x]*c1 >= 1000) continue;for(int y = 0; y < k; y++){if(a[y]*c1 >= 1000) continue;c2 = a[x]*10 + a[y];if(isok(c1*a[x]) && isok(c1*a[y]) && isok(c1*c2))ans++;}}}}}fout << ans << endl;}fout.close();return 0;}
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