HDU-4475 Downward paths(找规律)
Downward paths
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 844 Accepted Submission(s): 282
Thanks to LCY, I have this chance to share my ideas and works with you. Good luck and have fun!
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We have a graph with size = N like that in Figure 1. Then we are going to find a downward path from the top node to one bottom node.
First, we select the top node as the beginning. Then at any node, we can go horizontally or downward along the blue edge and reach the next node. The finding will be end when we reach one of the bottom nodes. After that we can get a downward path from the top node to one bottom node. Note that we can not pass a blue edge that we have passed ago during each finding.
Your task is to calculate there exists how many downward paths.
For each test case, there is only one integer N (1 <= N <= 10^18) indicates the size of the graph.
Because the answer may be very large, you should just output the remainder of it divided by 1000003.
1
2
8
For Sample 2, the yellow paths in Figure 2 show the 8 downward paths.
********************************************************************************************
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define mset(a, b) memset((a), (b), sizeof(a))
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = +;
const int mod = ;
LL a[maxn];
int main()
{
a[] = ;
for(int i=;i<=maxn;i++)
a[i] = (a[i-] * i * )%mod;
LL T;
cin >> T;
while(T--){
LL n;
cin >> n;
if(n >= mod ){
cout << "" << endl;
continue;
}
cout << a[n] << endl;
}
return ;
}
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