题目:找(A,B)区间内所有的数的二进制形式中0的个数比1的个数多的数字的数量。
变成二进制,dp[pos][num0][num1]表示进行到pos这个数位时0的个数为num0,1的个数为num1时的值
#include <cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll A,B,dp[34][34][34],wei[34];
ll dfs(int pos,int num0,int num1,int limit,int lead)
{
if(pos<1)
return num0>=num1;
if(!limit&&dp[pos][num0][num1]!=-1)
return dp[pos][num0][num1];
int up=limit?wei[pos]:1;
ll ans=0;
for(int i=0;i<=up;i++)
{
if(i==0)
{
if(lead)
ans+=dfs(pos-1,num0,num1,limit&&i==up,i==0&&lead);
else
ans+=dfs(pos-1,num0+1,num1,limit&&i==up,i==0&&lead);
}
else {
ans+=dfs(pos-1,num0,num1+1,limit&&i==up,i==0&&lead);
}
}
if(!limit) dp[pos][num0][num1]=ans;
return ans;
}
ll solve(ll x)
{
int len=0;
while(x)
{
wei[++len]=x%2;
x/=2;
}
ll ans=dfs(len,0,0,1,1);
return ans;
}
int main()
{
memset(dp,-1,sizeof dp);
while(~scanf("%lld%lld",&A,&B))
{
ll ans=solve(B)-solve(A-1);
printf("%lld\n",ans);
}
return 0;
}