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
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)
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).
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.
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
#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 ;
}
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