toj 4613 Number of Battlefields

时间限制(普通/Java):1000MS/3000MS 内存限制:65536KByte
总提交: 8 测试通过:6

描述

In the previous problem, we assume the perimeter of the figure equals to p, how many battlefields are possible? For example, there are no battlefields possible forp < 8, but two for p = 8:

Here are the nine battlefields for p=10:

You’re asked to output the number of battlefields modulo 987654321.

输入

There will be at most 25 test cases, each with a single integer p ( 1<=p<=109), the perimeter of the battlefield. The input is terminated by p = 0.

输出

For each test case, print a signle line, the number of battlefields, modulo 987654321.

样例输入

7
8
9
10
0

样例输出

0
2
0
9

/*
本题用快速矩阵幂的方法求斐波那契序列
#include <cstdio>
#include <iostream>using namespace std;const int MOD = 10000;int fast_mod(int n)    // 求 (t^n)%MOD
{int t[2][2] = {1, 1, 1, 0};int ans[2][2] = {1, 0, 0, 1};  // 初始化为单位矩阵int tmp[2][2];    //自始至终都作为矩阵乘法中的中间变量while(n){if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t{for(int i = 0; i < 2; ++i)for(int j = 0; j < 2; ++j)tmp[i][j] = ans[i][j];ans[0][0] = ans[1][1] = ans[0][1] = ans[1][0] = 0;  // 注意这里要都赋值成 0for(int i = 0; i < 2; ++i)    //  矩阵乘法{for(int j = 0; j < 2; ++j){for(int k = 0; k < 2; ++k)ans[i][j] = (ans[i][j] + tmp[i][k] * t[k][j]) % MOD;}}}//  下边要实现  t *= t 的操作，同样要先将t赋值给中间变量  tmp ，t清零，之后 t = tmp* tmpfor(int i = 0; i < 2; ++i)for(int j = 0; j < 2; ++j)tmp[i][j] = t[i][j];t[0][0] = t[1][1] = 0;t[0][1] = t[1][0] = 0;for(int i = 0; i < 2; ++i){for(int j = 0; j < 2; ++j){for(int k = 0; k < 2; ++k)t[i][j] = (t[i][j] + tmp[i][k] * tmp[k][j]) % MOD;}}n >>= 1;}return ans[0][1];
}int main()
{int n;while(scanf("%d", &n) && n != -1){printf("%d\n", fast_mod(n));}return 0;
}
*/#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod=987654321;ll A[2][2],B[2][2],T[2][2];void pow(int n)
{if(n==0){for(int i=0;i<2;i++)for(int j=0;j<2;j++)B[i][j]=(i==j);return;}if(n&1){pow(n-1);for(int i=0;i<2;i++)for(int j=0;j<2;j++){T[i][j]=0;for(int k=0;k<2;k++)T[i][j]=(T[i][j]+A[i][k]*B[k][j])%mod;}for(int i=0;i<2;i++)for(int j=0;j<2;j++){B[i][j]=T[i][j];}}else{pow(n/2);for(int i=0;i<2;i++)for(int j=0;j<2;j++){T[i][j]=0;for(int k=0;k<2;k++)T[i][j]=(T[i][j]+B[i][k]*B[k][j])%mod;}for(int i=0;i<2;i++)for(int j=0;j<2;j++){B[i][j]=T[i][j];}}
}int main()
{int n;A[0][0]=1;    A[0][1]=1;A[1][0]=1;    A[1][1]=0;while (scanf("%d", &n) == 1 && n){ll ans=0;if(n&1) ans=0;else if(n<4) ans=0;else{pow(n-4);//从第5个开始ans=B[0][0]-n/2+1;//B[0][0]求出对应p下的方块排列总数，（n/2-1）是相应的形成规则矩形的个数，两者之差即为所求的符合要求的方块个数ans%=mod;if(ans<0)ans+=mod;}printf("%lld\n",ans);}return 0;
}