hdu_4944_FSF’s game

Posted 2020-10-15 会飞的雅蠛蝶

FSF has programmed a game.
In this game, players need to divide a rectangle into several same squares.
The length and width of rectangles are integer, and of course the side length of squares are integer.

After division, players can get some coins.
If players successfully divide a AxB rectangle(length: A, width: B) into KxK squares(side length: K), they can get A*B/ gcd(A/K,B/K) gold coins.
In a level, you can’t get coins twice with same method.
(For example, You can get 6 coins from 2x2(A=2,B=2) rectangle. When K=1, A*B/gcd(A/K,B/K)=2; When K=2, A*B/gcd(A/K,B/K)=4; 2+4=6; )

There are N*(N+1)/2 levels in this game, and every level is an unique rectangle. (1x1 , 2x1, 2x2, 3x1, ..., Nx(N-1), NxN)

FSF has played this game for a long time, and he finally gets all the coins in the game.
Unfortunately ,he uses an UNSIGNED 32-BIT INTEGER variable to count the number of coins.
This variable may overflow.
We want to know what the variable will be.
(In other words, the number of coins mod 2^32)

InputThere are multiply test cases.

The first line contains an integer T(T<=500000), the number of test cases

Each of the next T lines contain an integer N(N<=500000).OutputOutput a single line for each test case.

For each test case, you should output "Case #C: ". first, where C indicates the case number and counts from 1.

Then output the answer, the value of that UNSIGNED 32-BIT INTEGER variable.Sample Input

3
1
3
100

Sample Output

Case #1: 1
Case #2: 30
Case #3: 15662489


        
 

HinIn the second test case, there are six levels(1x1,1x2,1x3,2x2,2x3,3x3)Here is the details for this game:

1x1: 1(K=1); 1x2: 2(K=1); 1x3: 3(K=1);  2x2: 2(K=1), 4(K=2); 2x3: 6(K=1); 3x3: 3(K=1), 9(K=3); 
1+2+3+2+4+6+3+9=30


根据题意写出式子，ans[n]=ans[n-1]+∑ n*i*k/gcd（n,i）

∑ n*i*k/gcd（n,i）  //gcd(n,i)/k 为n的因子,
=n?(1/a1+2/a2+?+n/an)   //gcd(n,i)/k =ai，ai为n的因子

也就是说每个n的值对应了n的所有因子的贡献值之和。
比如n=2,因子1,2.
1的贡献　　2　　4
2的贡献　　 　　2
设mi为ai的因子  ∑ n*i*k/gcd（n,i）=n*[（1*m1/m1+2*m1/m1+...n/m1）  +（1*m2/m2+2*m2/m2+...n/m2） +........+（1*mn/mn+2*mn/mn+...n/m）]   //主要想清楚k/gcd（）的值,k的变化对应mi的值
设sum（mi）=（1*mi/mi+2*mi/mi+...n/mi）=（n/m）*（n/m+1）/2
ans[n]=ans[n-1]+sum(mi)*mi



 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
typedef long long LL;
#include<algorithm>
using namespace std;
#define N 500005
const LL mod=1LL<<32;
LL ans[N];
LL num[N];
int main()
{
    //freopen("i.txt","r",stdin);
    //freopen("o.txt","w",stdout);
    for(int i=1;i<N;i++)
    {
        for(LL j=i;j<N;j+=i)
        {
            num[j]+=(j/i)*(j/i+1)/2;
            //cout<<j<<" "<<num[j]<<endl;
        }
    }
    ans[1]=1;
    for(LL i=2;i<=500000;i++)
    {
        ans[i]=ans[i-1]+num[i]*i%mod;
        ans[i]%=mod;
        //cout<<ans[i]<<endl;
    }
    int t;
    scanf("%d",&t);
    for(int l=1;l<=t;l++)
    {
        int n;
        scanf("%d",&n);
        cout<<"Case #"<<l<<": "<<ans[n]<<endl;

    }
}