NJUST 1739 - Count The Carries

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题目：NJUST 1739 - Count The Carries

思路：用一个30位左右的数组，记录一下从0~n中各位出现1的频率，然后递推累加上去，就搞定了

这里鄙视一下小胖（竟然说数据太大了搞不出来） = =

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <iostream>using namespace std;long long num[101];long long a,b;void solve(){    memset(num,0,sizeof(num));    long long p=2;    b++;    for(int i=1;i<=30;i++)    {        num[i]+=b/p*(p/2);        int remain=b%p;        if(remain<=p/2)            ;        else            num[i]+=remain-p/2;        p*=2;    }    p=2;    for(int i=1;i<=30;i++)    {        num[i]-=a/p*(p/2);        int remain=a%p;        if(remain<=p/2)            ;        else            num[i]-=remain-p/2;        p*=2;    }    long long sy=0;    for(int i=1;i<=100;i++)    {        sy+=num[i]/2;        num[i+1]+=num[i]/2;    }    printf("%lld\n",sy);    return ;}int main(){    while(scanf("%lld%lld",&a,&b)!=EOF)    {        solve();    }    return 0;}