loj1245(数学)

传送门：Harmonic Number (II)

题意：求sum=n/1+n/2+n/3+...+n/n。(n<2^31)

分析：在一定的区间内n/i的值是一定的，因此要跳过这段区间来加速求解。

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <limits.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define LL long long
#define mod 100000000
#define inf 0x3f3f3f3f
#define eps 1e-6
#define N 10000000
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PII pair<int,int>
using namespace std;
inline LL read()
{
    char ch=getchar();LL x=0,f=1;
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{
    int n;
    int T,cas=1;
    T=read();
    while(T--)
    {
       n=read();
       LL ans=0;
       for(LL i=1,last=0;i<=n;i=last+1)
       {
           last=n/(n/i);
           ans+=(last-i+1)*(n/i);
       }
       printf("Case %d: %lld
",cas++,ans);
    }
}