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  • HDOJ 4944 FSF’s game


    http://blog.csdn.net/keshuai19940722/article/details/38519681

    不明真相的补一发。。。

    FSF’s game

    Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 448    Accepted Submission(s): 215


    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
    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
     




    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long int LL;
    
    const int maxn=500050;
    const LL mod=1LL<<32;
    
    LL f[maxn],s[maxn];
    
    void init()
    {
        for(int i=1;i<maxn;i++)
        {
            for(int k=1;k*i<maxn;k++)
            {
                f[i*k]+=(1LL+k)*k/2;
                f[i*k]%=mod;
            }
            f[i]=f[i]*i;
            f[i]%=mod;
        }
        for(int i=1;i<maxn;i++)
        {
            s[i]=(f[i]+s[i-1])%mod;
        }
    }
    
    int main()
    {
        init();
        int T_T,cas=1;
        scanf("%d",&T_T);
        while(T_T--)
        {
            int x;
            scanf("%d",&x);
            printf("Case #%d: %lld
    ",cas++,s[x]);
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/mthoutai/p/6936201.html
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