hdu 4944

FSF’s game

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1135    Accepted Submission(s): 582

Problem Description

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, 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)

 

Input

There 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).

 

Output

Output 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

 

 

析：指定n阶矩形共n*(n+1)/2个，为边长A*B为(1*1，1*2，2*2，2*3，...，(n-1)*n，n*n)，然后计算将这n阶矩形分解成k*k的正方形代价为A*B/(gcd(A/k，B/k))，要求计算总代价

　　可令代价和为res[n] ，则知res[n-1]包括(1*1，1*2，2*2，2*3，...，(n-2)*(n-1)，(n-1)*(n-1))，而res[n] = res[n-1] + 代价(1*n，2*n，3*n，...，(n-1)*n，n*n)，

　　故可令tmp[n] = 代价(1*n，2*n，3*n，...，(n-1)*n，n*n)，所以求总代价，先预处理所有tmp即可。而对每一个矩形i*n的代价 = i*n/gcd(i/k，n/k) = n*(i/j1+i/j2+i/j3...) (其中gx为n与i的公因数)

　　所以可以枚举所有i与n的公因子，对每一个公因子j，存在于n以内的数有j，2*j，3*j，...，n/j，故代价i*n = Σn*(1+2+...+n/j) = Σn*(1+n/j)*(n/j)/2

 

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ll long long
#define mod 4294967296
using namespace std;
const int N = 500005;
int t, n, m, ans;
ll res[N], tmp[N];
void init(){
    for(ll i = 1; i < N; i ++){
        for(ll k = i; k < N; k += i){
            tmp[k] += (1+k/i)*(k/i)/2;
            tmp[k] %= mod;
        }
    }
    for(int i = 1; i < N; i ++){
        res[i] = (res[i-1]+tmp[i]*i)%mod;
    }
}
int main(){
    init();
    scanf("%d", &t);
    for(int i = 1; i <= t; i ++) {
        scanf("%d", &n);
        printf("Case #%d: %lld
", i, res[n]);
    }
    return 0;
}