POJ-3252 Avenger

Posted 2020-10-01 gggyt

　　题意：在区间中，他们化成2进制的数的0的个数大于等于1的数有多少个。

　　思路：我们需要记录上一次0和1的个数，此外我们还要特别注意一下前导0。

　　如果前面全是0的时候我们就要注意下一位是不是还是0，如果一直都是0那么这个数也满足条件。

/*  gyt
       Live up to every day            */
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>3
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 100+10;
const ll maxm = 1e7;
const int mod = 1000000007;
const int INF = 1<<30;
const db eps = 1e-9;
const ll Max=1e19;
int a[maxn];
ll dp[maxn][maxn][maxn];

ll dfs(int pos, int num0, int num1, int limit, int ncan) {
    if (pos==-1) {
        if (ncan)  return 1;
        if (num0>=num1)  return 1;
        return 0;
    }
    if (!limit && !ncan && dp[pos][num0][num1]!=-1)  return dp[pos][num0][num1];
    int up=limit?a[pos]:1;
    ll tmp=0;
    for (int i=0; i<=up; i++) {
        if (ncan) {
            if (i==0)
                tmp+=dfs(pos-1, 0, 0, limit&&i==a[pos], 1);
            else
                tmp+=dfs(pos-1, 0, 1, limit&&i==a[pos], 0);
        }
        else {
            if (i==0)
                tmp+=dfs(pos-1, num0+1, num1, limit&&i==a[pos], 0);
            else
                tmp+=dfs(pos-1, num0, num1+1, limit&&i==a[pos], 0);
        }
    }
    if (!limit)  dp[pos][num0][num1]=tmp;
    return tmp;
}
ll deal(ll x) {
    int pos=0;
    while(x) {
        a[pos++]=x%2;
        x/=2;
    }
    dfs(pos-1, 0, 0, 1, 1);
}
void solve() {
    ll n, m;  scanf("%lld%lld", &n, &m);
    memset(dp, -1, sizeof(dp));
    ll a=deal(m), b=deal(n-1);
    //cout<<a<<" "<<b<<endl;
    printf("%lld\n", a-b);
}
int main() {
    int t=1;
    //freopen("in.txt", "r", stdin);
    //scanf("%d", &t);
    for (int T=1; T<=t; T++) {
        solve();
    }
}