POJ 1426 Find the Multiple(二维DP)
总结
1. 编程之美讲了剪枝解法, 这里给出动规解法
2. dp[i][j]%n = j. dp[i][j] 是 mod n 余 j 的最小值
代码
/* * source.cpp * * Created on: Apr 6, 2014 * Author: sangs */ #include <stdio.h> #include <iostream> #include <string> #include <vector> #include <memory.h> using namespace std; typedef unsigned long long LL; LL dp[200][500]; LL cal(int n) { LL lastModed = 1; LL expr10 = 1; memset(dp, 0, sizeof(dp)); dp[0][1] = 1; for(int i = 1; ; i ++) { for(int j = 1; j < n; j ++) { dp[i][j] = dp[i-1][j]; } LL thisModed = (lastModed*(10%n))%n; lastModed = thisModed; expr10 *= 10; if(dp[i-1][thisModed] == 0) { dp[i][thisModed] = expr10; } //cout << dp[i][thisModed] << endl; for(int j = 1; j < n; j ++) { if(dp[i-1][j] == 0) { continue; } LL candy = (j + thisModed)%n; if(dp[i][candy] == 0) { dp[i][candy] = expr10 + dp[i-1][j]; } } if(dp[i][0] != 0) { return dp[i][0]; } } return 1; } int main() { freopen("input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF && n != 0) { LL res = cal(n); printf("%lld\n", res); } return 0; }