POJ3252Round Numbers(数位DP)

题意：算出区间内二进制中0的个数大于等于1的个数的数字有多少个

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;#define N 50int dp[N][N][N], bit[N];int dfs(int pos, int num0, int num1, bool pre, bool limit){    if(pos==-1)    {        return num0>=num1;    }    if(!limit&&!pre&&dp[pos][num0][num1]!=-1) return dp[pos][num0][num1];    int up=limit?bit[pos]:1;    int ret=0;    for(int i=0; i<=up; i++)    {        if(i==0)        {            if(pre) ret+=dfs(pos-1, num0, num1, pre, limit&&i==up);            else ret+=dfs(pos-1, num0+1, num1, pre, limit&&i==up);        }        else            ret+=dfs(pos-1, num0, num1+1, false, limit&&i==up);    }    if(!limit&&!pre) dp[pos][num0][num1]=ret;    return ret;}int solve(int x){    int len=0;    while(x)    {        bit[len++]=x%2;        x>>=1;    }    return dfs(len-1, 0, 0, true, true);}int main(){    int L, R;    memset(dp, -1, sizeof(dp));    while(cin>>L>>R)    {        printf("%d\n", solve(R)-solve(L-1));    }    return 0;}