USACO 1.3 Prime Cryptarithm

开始时用数组对给出的数字加以标记，然后三位数 111--999 两位数11--99分别相乘加以判断。

/*ID: acmerfi1PROG: crypt1LANG: C++*/#include <stdio.h>#include <string.h>#include <stdlib.h>#define MAX 12int N;int hash[12], num, sum = 0;int exsit(int number) //判断是否在给定的数字中 {while(number){if(hash[number%10]) number /= 10;else return 0;}return 1;}int countDigit(int i, int j) // 计算乘积的位数 {int sum = i * j;int count = 0;if(exsit(i) && exsit(j) && exsit(sum)){while(sum){count++;sum /= 10;}}else return 0;return count;}int judge(int i, int j) // 判断是否满足题目中的条件 {if(4 == countDigit(i, j) && 3 == countDigit(i, j%10) && 3 == countDigit(i, j/10)){return 1;}else return 0;}int main(){freopen("crypt1.in", "r", stdin);freopen("crypt1.out", "w", stdout);memset(hash, 0, sizeof(num));scanf("%d", &N);for(int i = 0; i < N; i++){scanf("%d", &num);hash[num] = 1;}for(int i = 111; i <= 999; i++){for(int j = 11; j <= 99; j++){if(judge(i,j)) sum++;}}printf("%d\n", sum);return 0;}