HDU 4944 FSF’s game 一道好题

FSF’s game

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 727    Accepted Submission(s):
377

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

 

 

题意：略。

思路：对于A*B/gcd(A/k,B/k) 看成 N*x/a ,其中x未知，N已知，a是N的因子。

　　　（因为a必然是N的因子）

        1.现在我们这样转化后，就开始一个一个枚举a了。（我们把a看成了gcd()的整体来看。）

        2.对于一个确定的a值，假设为ai，那么我们现在要做的就是找出 （N*x/a ）满足要求的x来。

          并对它进行求和sum（xi/a）*N；（因为N始终没有变化呀。）

          此时 a = gcd(N/k,x/k) 可以转化成  gcd(N,x) = k*a, 

          那么，对于x的取值范围我们知道，是[1，N]，求gcd（N，x）=k*a （k是>0的正整数）

          其实就是在[1,N]里，a的倍数，a , 2a , 3a ,4a,,,,,N/a*a , 正确吗？

          会不会遗漏，gcd（）=k*a，就是代表最大公约数是a的倍数。

　　　 这样的话，我们就对x进行求和了。sum = a(1+2...N/a) = a*(1+N/a)*N/a/2 =>(1+N/a)*N/2;

         最后根据式子A*x/a,那么就变成  (1+N/a)*N/a /2 * N;

       筛选，dp即可。

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;
 typedef __int64 LL;

 const int maxn = 5e5+;
 LL p = ;
 LL dp[maxn];
 void init()
 {
     int j,tmp;
     for(j=;j<=;j++)p=p*;

     for(int i=;i<maxn;i++){
         tmp = i;
         for(j=;(tmp=i*j)<maxn;j++){
             dp[tmp]=(dp[tmp]+((LL)(+j)*(LL)j)/)%p;
         }
     }
     dp[]=;
     for(int i=;i<maxn;i++){
         dp[i]=(dp[i-]+dp[i]*i)%p;
     }
 }
 int main()
 {
     int T,n;
     init();
     scanf("%d",&T);
     for(int t=;t<=T;t++)
     {
         scanf("%d",&n);
         printf("Case #%d: %I64d\n",t,dp[n]);
     }
     return ;
 }