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
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.
#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]);
}
}
表要是表达式的转化很巧妙。当推理出 f(n) = Σ n * (i/c1 + i/c2 + ... + i/cm), cj为gcd(i, n)的所有因子时,我们枚举因子cj。相当于横向的式子先纵向统计了。