nuc2016第一次月赛A

最新推荐文章于 2021-05-16 17:35:49 发布

原创最新推荐文章于 2021-05-16 17:35:49 发布 · 333 阅读

0 ·

CC 4.0 BY-SA版权

本文探讨了郭杰女士从原点出发，在不重复经过已访问点的情况下，行走n步的不同路径数量计算方法。通过递推算法实现了路径计数，并采用模运算确保数值大小在限定范围内。

描述

The first lady of our lab is GuoJie, GuoJie likes walking very much.

Today GuoJie walks from the original point (0, 0), everytime he(may be she?) can go up or left or right a step.

But she can't go back the point where she have visited.

For example, if he goes up a step, she will be at (1, 0) and she never comes back the point.

Now, if she can walk n(n <= 100000) steps, can you find how many ways she can walk? the result mod 1e9 + 7.

输入

There will be T (T <= 100) cases, each case will input a n.

输出

For each group of input integers you should output how many ways GuoJie can walk in one line, and with one line of output for each line in input.

样例输入

1
2

样例输出

7#include <iostream>
#include<stdio.h>
using namespace std;
long long num[1000006];
void gon()
{
    long long a=1;
    long long  b=0;
    num[0]=1;
    for(int i=1;i<1000000;i++)
    {
        a=a+b;
        b=num[i-1]-a;
        num[i]=3*a+2*b;
        while(num[i]>=1000000007)
            num[i]-=1000000007;
    }
    num[0]=0;
    return ;
}

int main()
{
    int t,n;
    gon();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n",num[n]);
    }
    return 0;
}