【数位DP】Round Numbers POJ

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Think:
1知识点：数位DP+二进制
2题意：输入一个区间判断有多少个“Round Number”，“Round Number”的定义为其二进制表示中0的数量大于等于1的数量

vjudge题目链接

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int dp[40][40][40], tp, link[40];int dfs(int pos, int cnt0, int cnt1, bool lead, bool is_max);int solve(int x);int main(){    memset(dp, -1, sizeof(dp));    int l, r;    while(~scanf("%d %d", &l, &r)){        printf("%d\n", solve(r)-solve(l-1));    }    return 0;}int solve(int x){    tp = 0;    while(x){        link[tp++] = x%2;        x >>= 1;    }    return dfs(tp-1, 0, 0, true, true);}int dfs(int pos, int cnt0, int cnt1, bool lead, bool is_max){    if(pos == -1)        return cnt0 >= cnt1 && !lead;    if(!is_max && !lead && ~dp[pos][cnt0][cnt1])        return dp[pos][cnt0][cnt1];    int up_top = 1;    if(is_max)        up_top = link[pos];    int sum = 0;    for(int i = 0; i <= up_top; i++){        if(lead){            if(i == 0){                sum += dfs(pos-1, 0, 0, true, false);            }            else if(i == 1){                sum += dfs(pos-1, 0, 1, false, is_max && i == up_top);            }        }        else {            if(i == 0){                sum += dfs(pos-1, cnt0+1, cnt1, false, is_max && i == up_top);            }            else if(i == 1){                sum += dfs(pos-1, cnt0, cnt1+1, false, is_max && i == up_top);            }        }    }    if(!is_max)        dp[pos][cnt0][cnt1] = sum;    return sum;}

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