POJ3252Round Numbers(数位dp)

题意

给出区间$[a, b]$，求出区间内的数转成二进制后$0$比$1$多的数的个数

$1 \leqslant a, b \leqslant 2,000,000,000$

sol

比较zz的数位dp

直接在二进制下dp就好

$f[i][ze][on]$表示第$i$位，填了$ze$个$0$，$on$个1的方案数

#include<cstdio>
#include<cstring>
#include<iostream>
// #include<map>
using namespace std;
#define ll long long
const ll maxn = 101;
ll a, b;
ll num[maxn], tot, f[maxn][maxn][maxn];
ll dfs(ll x, bool lim, ll ze, ll on) {
    if(x == 0) return 
        (ze != -1) && (on != -1) && (ze >= on);
    if(!lim && f[x][ze][on]) return f[x][ze][on];
    ll ans = 0;
    for(ll i = 0; i <= (lim ? num[x] : 1); i++) {
        if(i == 0) ans += dfs(x - 1, lim && (i == num[x]), ze == -1 ? 1 : ze + 1, on);
        else {
            if(on == -1) ans += dfs(x - 1, lim && (i == num[x]), 0, 1);
            else ans += dfs(x - 1, lim && (i == num[x]), ze, on + 1);
        }
    }
    if(!lim) f[x][ze][on] = ans;
    return ans;
}
ll solve(ll x) {
    tot = 0;
    while(x) num[++tot] = x % 2, x >>= 1;
    return dfs(tot, 1, -1, -1);
}
int main() {
    cin >> a >> b;
    cout << solve(b) - solve(a - 1);
    return 0;
}
/*
1234 4444
2
*/