POJ 3252 Round Numbers （二进制数位dp）

题意:

求[l,r]区间里满足转化成二进制后“0”个数比“1”多的数的个数

分析:

破题想了2天,老想着填0→9,想了几个状态都没办法,都跟暴力没啥区别
瞄到二进制的时候−−尼玛原来这是个水题−−只要填0→1就好了
思维定势啊, 思维局限啊！！

dp[i][m0][m1]:=i位,二进制0,1的个数,注意这里要忽略前导0哟

代码:

////  Created by TaoSama on 2015-10-21//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int l, r;int dp[35][35][35], digit[35];int dfs(int i, int m0, int m1, bool first, bool e) {    if(!i) return m0 >= m1;    if(!e && ~dp[i][m0][m1]) return dp[i][m0][m1];    int ret = 0;    int to = e ? digit[i] : 1;    for(int d = 0; d <= to; ++d)        ret += dfs(i - 1, first && !d ? i == 1 : m0 + (d ^ 1),                   m1 + d, first && !d, e && d == to);    if(!e) dp[i][m0][m1] = ret;    return ret;}int calc(int x) {    int cnt = 0;    for(; x; x >>= 1) digit[++cnt] = x & 1;    return dfs(cnt, 0, 0, 1, 1);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    memset(dp, -1, sizeof dp);    while(scanf("%d%d", &l, &r) == 2) {        printf("%d\n", calc(r) - calc(l - 1));    }    return 0;}