- 512K
Description:
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// Q.cpp
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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.
sample input
1 233 1 666
Sample output
234
667
This question directly pastes the code it gives will cause the program memory overrun error, so you need to read the code it gives, the code seeks the n-term sequence, from 1 to the sum of the sum of any i-term combination after cumulative multiplication , for example, there are 3 items, the answer is 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 <bits/stdc++.h>
using namespace std;
typedef unsigned long long unll;
typedef long long ll;
const int m = 1e9 + 7;
intmain()
{
int n;
ll a;
ll ans;
while(scanf("%d", &n) != EOF){
years = 1;
for(int i = 0; i < n; i ++){
scanf("%lld", &a);
years = years * (a + 1) % m;
//cout << a[i] <<endl;
}
//cout << n << endl;
printf("%lld\n", ans);
}
return 0;
}