URAL - 1057 Amount of Degrees--数位dp

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原题链接：http://vjudge.net/problem/URAL-1057

题意：[ x , y ]里一共有多少个数可以由k个b整数幂组成。

分析：http://wenku.baidu.com/link?url=q4atTAoZVGlV6sfo0fhED06ogbktY38_TZkGWLkuOpTRiqyI-eDyarkTeL10fv2GdUe53DMIloZ_sD0gZF6xK1ljbcJH1NlLgdyh4aVcGXi

#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#include<string>#include<stdio.h>#define INF 99999999#define eps 0.0001using namespace std;int dp[35][35];//dp[i][j]表示深度为i有j个1的种类,根深度为0int x, y, k, b;int cnt[35];void init(){for (int i = 0; i <= 31; i++){dp[i][0] = 1;for (int j = 1; j <= i; j++)dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];}}int solve(int n){int len = 0;int ans = 0;int ik = k;int ib = b;while (n){cnt[len++] = n%ib;n /= ib;}for (int i = len - 1; i >= 0; i--){if (cnt[i] == 1){ans += dp[i][ik];//以0开头，找到k个1；然后k--，固定开头为1，继续做ik--;}else if (cnt[i] > 1){ans += dp[i + 1][ik];break;}if (ik < 0)//注意是小于0return ans;}if (ik == 0)//n本身满足ans++;return ans;}int main() {init();while (~scanf("%d%d%d%d", &x, &y, &k, &b)){printf("%d\n", solve(y) - solve(x - 1));}return 0;}

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