poj 3252 Round Numbers

题目链接：Round Numbers

题目大意：给你一个区间，求这个区间里面数转化成二进制后0的个数大于1的个数有多少个

题目思路：直接数位dp，不过在考虑的时候需要考虑这样一种情况，如果有前导0的情况我们需要用一个bool去判断一下有没有前导零的存在，然后去比较0和1的个数差就好了

#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;int dp[205][205][205],a[205],n,m;int dfs(int pos,int num0,int num1,bool lead,bool limit){    if(pos == -1){        if(lead == 0) return 1;        if(num0 >= num1) return 1;        return 0;    }    if(!limit&&lead&&dp[pos][num0][num1] != -1) return dp[pos][num0][num1];    int up = limit?a[pos]:1;    int tmp = 0;    for(int i = 0;i <= up;i++){        if(!lead){            if(i == 0) tmp += dfs(pos-1,0,0,false,limit&&i == up);            else tmp += dfs(pos-1,0,1,true,limit&&i == up);        }        else{            if(i == 0) tmp += dfs(pos-1,num0+1,num1,true,limit&&i == up);            else tmp += dfs(pos-1,num0,num1+1,true,limit&&i == up);        }    }    if(!limit&&lead) dp[pos][num0][num1] = tmp;    return tmp;}int solve(int x){    int pos = 0;    while(x){        a[pos++] = x%2;        x/=2;    }    return dfs(pos-1,0,0,false,true);}int main(){    memset(dp,-1,sizeof(dp));    while(~scanf("%d%d",&n,&m)){        printf("%d\n",solve(m)-solve(n-1));    }    return 0;}