POJ 3252 Round Numbers 数位DP

最新推荐文章于 2021-05-28 22:57:56 发布

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最新推荐文章于 2021-05-28 22:57:56 发布

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原文链接：http://www.cnblogs.com/rolight/p/3886680.html

本文介绍了一种使用动态规划方法来计算两个给定数值范围内特定二进制子序列出现次数的算法。该算法首先将给定的整数转换为二进制形式，并通过递归函数遍历所有可能的二进制位组合来统计符合要求的子序列数量。

注意只有高位放了1之后才能开始统计

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>

using namespace std;

typedef long long LL;

int lim[256],len = 0;
LL f[40][40][40];

void getlim(LL N) {
    memset(lim,0,sizeof(lim));
    memset(f,-1,sizeof(f));
    len = 0;
    LL tmp = N;
    while(tmp) {
        lim[len++] = tmp & 1;
        tmp >>= 1;
    }
}

LL dfs(int now,int cz,int co,int bound,int first) {
    LL &note = f[now][cz][co];
    if(now == 0) {
        if(first) return cz >= co;
        else return 0;
    }
    if(!bound && note != -1 && first) {
        return note;
    }
    int m = bound ? lim[now - 1] : 1;
    LL ret = 0;
    for(int i = 0;i <= m;i++) {
        if(first) {
            ret += dfs(now - 1,cz + !i,co + i,bound && i == lim[now - 1],first);
        } else {
            if(i == 0) ret += dfs(now - 1,cz,co,0,0);
            else ret += dfs(now - 1,0,1,bound && i == lim[now - 1],1);
        }
    }
    if(!bound && first) note = ret;
    return ret;
}

int main() {
    LL N,M;
    while(cin >> N >> M) {
        LL ret1,ret2;
        getlim(M); ret1 = dfs(len,0,0,1,0);
        getlim(N - 1); ret2 = dfs(len,0,0,1,0);
        cout << ret1 - ret2 << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/rolight/p/3886680.html