2018 ACM-ICPC 中国大学生程序设计竞赛线上赛

•  512K

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 integers a[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

这题直接贴它给的代码会出现程序内存超限的错误，所以要读懂它给出的代码，代码求的是n项序列，从1加到任意i项组合累乘之后之和的和，比如有3项，则答案为1+a[1] + a[2] + a[3] + a[1]*a[2]+a[1]*a[3]+a[2]*a[3] + a[1]*a[2]*a[3]

#include 

using namespace std;
typedef unsigned long long unll;
typedef  long long ll;
const int m = 1e9 + 7;
int main()
{
    int n;
    ll a;
    ll ans;
    while(scanf("%d", &n) != EOF){
        ans = 1;
        for(int i = 0; i < n; i ++){
            scanf("%lld", &a);
            ans = ans * (a + 1) % m;
            //cout << a[i] <