lightoj 1032 二进制的dp

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

const int maxn = 33;
const int INF  = 0x3f3f3f;

long long  dp[maxn];
int sum[9] = {0,0,0,1,1,1,2,4,4};
long long int N;

inline long long res(int i){
return (1<<i) + (1<<(i-1));
}

int main()
{
freopen("E:\\acm\\input.txt","r",stdin);
dp[2] = 1;
for(int i=3;i<=31;i++){
dp[i] = 2*dp[i-1] + (1<<(i-2));
}
int T;
cin>>T;
for(int cas=1;cas<=T;cas++){
cin>>N;
printf("Case %d: ",cas);
if(N <= 8){
printf("%d\n",sum
);
continue;
}
long long ans = 0;
for(int i=30;i>=3;i--){
long long temp = 1<<i;
if(N >= temp){
ans += dp[i];
long long diff = N - res(i);
if(diff >= 0){  //这个地方没有加等号WA了两次。
ans += diff + 1;
}
N -= temp;
}
if(N == 0) break;
}
cout<<ans+sum
<<endl;
}
}