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 = ;
LL A, B;
LL num[MAXN], tot, f[MAXN][MAXN][MAXN];
LL dfs(LL x, bool lim, LL ze, LL on) {
    if(x == ) return
        (ze != -) && (on != -) && (ze >= on);
    if(!lim && f[x][ze][on]) return f[x][ze][on];
    LL ans = ;
    for(LL i = ; i <= (lim ? num[x] : ); i++) {
        if(i == ) ans += dfs(x - , lim && (i == num[x]), ze == - ?  : ze + , on);
        else {
            if(on == -) ans += dfs(x - , lim && (i == num[x]), , );
            else ans += dfs(x - , lim && (i == num[x]), ze, on + );
        }
    }
    if(!lim) f[x][ze][on] = ans;
    return ans;
}
LL solve(LL x) {
    tot = ;
    while(x) num[++tot] = x % , x >>= ;
    return dfs(tot, , -, -);
}
int main() {
    cin >> A >> B;
    cout << solve(B) - solve(A - );
    return ;
}
/*
1234 4444
2
*/