POJ 3252 Round Numbers（数位dp）

http://poj.org/problem?id=3252

求区间[l,r]中二进制0的个数不小于1的个数的数的数目。

简单数位dp。

#include<cstdio>#include<cstring>#include<string>#include<cctype>#include<iostream>#include<set>#include<map>#include<cmath>#include<vector>#include<stack>#include<queue>#include<algorithm>using namespace std;int dp[35][70];int a[35];int dfs(int pos, int sta, bool lead, bool limit) {   if(pos == -1) return sta >= 32;   if(!lead && !limit && dp[pos][sta] >= 0) return dp[pos][sta];   int up = limit ? a[pos] : 1, ans = 0;   for(int i = 0; i <= up; i++) {      if(lead && i == 0) ans += dfs(pos-1, sta, true, limit && i == a[pos]);      else {         ans += dfs(pos-1, sta+(i==0?1:-1), false, limit && i == a[pos]);      }   }   if(!lead && !limit) dp[pos][sta] = ans;   return ans;}int solve(int x){   int pos = 0;   while(x) {      a[pos++] = x&1;      x >>= 1;   }   return dfs(pos-1, 32, true, true);}int main(){    int l, r;    memset(dp, -1, sizeof dp);    scanf("%d%d", &l, &r);    printf("%d\n", solve(r)-solve(l-1));}