Count The Carries - HDU 4588 水题

Count The Carries

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1158    Accepted Submission(s): 383

Problem Description

One day, Implus gets interested in binary addition and binary carry. He will transfer all decimal digits to binary digits to make the addition. Not as clever as Gauss, to make the addition from a to b, he will add them one by one from a to b in order. For example, from 1 to 3 (decimal digit), he will firstly calculate 01 (1)+10 (2), get 11，then calculate 11+11 (3)，lastly 110 (binary digit), we can find that in the total process, only 2 binary carries happen. He wants to find out that quickly. Given a and b in decimal, we transfer into binary digits and use Implus's addition algorithm, how many carries are there？

 

Input

Two integers a, b（0<=a<=b<1000000000), about 100000 cases, end with EOF.

 

Output

One answer per line.

 

Sample Input

1 21 31 41 6

 

Sample Output

0236

 

题意：从l加到r二进制进位一共发生多少次。

思路：预算出每个二进制位上共有多少个1，假设有n个1，那么这个位上的进位会发生n/2次。

AC代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;ll L,R;int T,t,n,m;ll num[110],POW[45];ll solve(ll ret,ll p){    int i,j;    ll k,ans;    ret++;    k=ret/p;    ans=k*p/2;    ret=ret-k*p;;    if(ret>p/2)      ans+=ret-p/2;    return ans;}int main(){    int i,j,k;    ll ans,ret;    POW[1]=2;    for(i=2;i<=40;i++)       POW[i]=POW[i-1]*2;    while(~scanf("%I64d%I64d",&L,&R))    {        memset(num,0,sizeof(num));        for(i=1;i<=40;i++)           num[i]=solve(R,POW[i])-solve(L-1,POW[i]);        ans=0;        for(i=1;i<=100;i++)        {            ret=num[i]/2;            num[i+1]+=ret;            ans+=ret;        }        printf("%I64d\n",ans);    }}