HDU 4944 FSF’s game 解題報告（遞推）

FSF’s game

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 72    Accepted Submission(s): 30

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, 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)

 

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

Hint

In 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

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

    解題報告： 比賽時沒有想出來。賽後看了神牛們的解體報告。給個鏈接：http://blog.csdn.net/a601025382s/article/details/38517329

    上篇說的很詳細了。爲什麼比賽時沒有搞出來纔是問題……

    貼個代碼：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iomanip>
using namespace std;
#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
#define bit(n) (1LL<<(n))
typedef long long LL;
typedef unsigned long long ULL;
void work();
int main()
{
#ifdef ACM
    freopen("input.in", "r", stdin);
#endif // ACM

    work();
}

/***************************************************/

const int maxn = 500000;
unsigned num[maxn + 5];
unsigned dp[maxn + 5];

void init()
{
    fff(i, 1, maxn)
        for(ULL j = i; j <= maxn; j += i)
            num[j] += (j/i+1)*j/i/2;

    fff(i, 1, maxn)
        dp[i] = dp[i-1] + num[i] * (unsigned)i;
}

void work()
{
    init();

    int T;
    scanf("%d", &T);
    fff(cas, 1, T)
    {
        int n;
        scanf("%d", &n);
        printf("Case #%d: %u\n", cas, dp[n]);
    }
}

    表要是表達式的轉化很巧妙。當推理出 f(n) = Σ n * (i/c1 + i/c2 + ... + i/cm), cj爲gcd(i, n)的所有因子時，我們枚舉因子cj。相當於橫向的式子先縱向統計了。