USACO section 1.3.4 Prime Cryptarithm
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1. 枚举，注意里边的几个剪枝，标在代码中了
2. 我的代码：
/*ID: dollar4PROG: crypt1LANG: C++*/#include <iostream>#include <fstream>#include <string>#include <algorithm>#include <cstring>using namespace std;int num[10], n;bool cmp(int a, int b){ return a < b;}bool findin(int a){ for (int i = 0; i < n; i++) if (num[i] == a) return true; return false;}int main(){ ofstream fout ("crypt1.out"); ifstream fin ("crypt1.in"); fin >> n; int i, j, k, l, m, s1, s2, s3, s4, s5; int cnt = 0, mm, nn, mn; for (i = 0; i < n; i++) fin >> num[i]; sort(num, num + n, cmp); for (i = 0; i < n; i++) for (j = 0; j < n; j++) for (k = 0; k < n; k++) for (l = 0; l < n; l++) for (m = 0; m < n; m++) { s1 = num[i]; s2 = num[j]; s3 = num[k]; s4 = num[l]; s5 = num[m]; mm = s4 * (s1 * 100 + s2 * 10 + s3); nn = s5 * (s1 * 100 + s2 * 10 + s3); mn = nn + 10 * mm; if (s1 * s4 > 9 || s5 * s1 > 9)//相乘必须是3位数 continue; //想加的约束 if (s1 * s4 + s2 * s4 / 10 >= 10 || s1 * s5 + s2 * s5 / 10 >= 10)//nn / 100 + mm % 100 / 10 < 10 || continue; //运算得到的结果有没有超出所给的数组之外的 if ( !findin(nn / 100) || !findin(nn % 100 / 10) || !findin(nn % 10) || !findin(mm / 100) || !findin(mm % 100 / 10) || !findin(mm % 10)) continue; if (!findin(mn / 1000) || !findin(mn % 1000 / 100) || !findin(mn % 100 / 10) || !findin(mn % 10)) continue; cnt++; } fout << cnt << endl; return 0;}2. 官方代码：The constraints of this problem are small enough that we can just try all possible products of 3 digit * 2 digit numbers, and look to see if all the correct digits are used.
The function "isgood" checks that a number is composed only of acceptable digits, and "isgoodprod" checks that all the lines of the multiplication are composed of acceptable digits.
#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h>int isgooddigit[10];/* isgooddigit[d] is set if d is an acceptable digit*//* check that every decimal digit in "n" is a good digit, and that it has the right number "d" of digits. */intisgood(int n, int d){ if(n == 0)return 0; while(n) {if(!isgooddigit[n%10]) return 0;n /= 10; d--; } if( d == 0 ) return 1; else return 0;}/* check that every product line in n * m is an okay number */intisgoodprod(int n, int m){ if(!isgood(n,3) || !isgood(m,2) || !isgood(n*m,4))return 0; while(m) {if(!isgood(n*(m%10),3)) return 0;m /= 10; } return 1;}voidmain(void){ int i, j, n, nfound; FILE *fin, *fout; fin = fopen("crypt1.in", "r"); fout = fopen("crypt1.out", "w"); assert(fin != NULL && fout != NULL); for(i=0; i<10; i++) { isgooddigit[i] = 0; } fscanf(fin, "%d", &n); for(i=0; i<n; i++) {fscanf(fin, "%d", &j);isgooddigit[j] = 1; } nfound = 0; for(i=100; i<1000; i++)for(j=10; j<100; j++) if(isgoodprod(i, j))nfound++; fprintf(fout, "%d\n", nfound); exit(0);}
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