（递归转递推）Copy and Submit II

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Description:

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// Q.cpp

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#include 

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using namespace std;

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const long long M = 1000000007;

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const long long MAXL = 1000000;

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long long a[MAXL];

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long long Q(int n, long long t)

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{

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    if(n < 0) return t;

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    return (Q(n - 1, t) + Q(n - 1, (t * a[n]) % M)) % M;

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}

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int main()

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{

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    int n;

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    while(cin >> n)

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    {

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        for(int i = 0; i < n; ++i)

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        {

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            cin >> a[i];

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            a[i] %= M;

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        }

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        cout << Q(n - 1, 1) << endl;

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    }

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    return 0;

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}

Input:

Input consists of several test cases. Each test case begins with an integer n. Then it's followed by n integersa[i]. 

0<n<=1000000

0<=a[i]<=10000

There are 100 test cases at most. The size of input file is less than 48MB. 

Output:

Maybe you can just copy and submit. Maybe not. 

样例输入

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233
1
666

样例输出

234
667

题目来源

2018 ACM-ICPC 中国大学生程序设计竞赛线上赛

比赛时看到这个代码，没想到递归转成递推，一直想着是不是有什么规律，，，道行太浅，继续修炼！

对了，不要用cin，cout,即使解除了iostream与stdio输出流的绑定，还是超时。

#include
#include
using namespace std;
const long long M=1000000007;
const long long MAXL=1000000;
long long a;

int main()
{
    int n;
    while(~scanf("%d",&n)){
        long long res=1;
        for(int i=0;i