Description:
// Q.cpp
using namespace std;
const long long M = 1000000007;
const long long MAXL = 1000000;
long long a[MAXL];
long long Q(int n, long long t)
{
if(n < 0) return t;
return (Q(n - 1, t) + Q(n - 1, (t * a[n]) % M)) % M;
}
int main()
{
int n;
while(cin >> n)
{
for(int i = 0; i < n; ++i)
{
cin >> a[i];
a[i] %= M;
}
cout << Q(n - 1, 1) << endl;
}
return 0;
}
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
1 233 1 666
样例输出
234 667
题意:其实题意就是 输入一数n,下面有n个数, 给你代码递归的意思就是,从这n 个数中取出 i (0<=i<=n) 相乘,输出所有可能得出的结果 相加的和;( 定义:从n个数取出0个相乘得时为1) 例:
n = 3; a[1] a[2] a[3]
输出 : 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];
上述那个递归的过程就是这个意思;
怎么才能简化这个过程呢,下面我们利用 数学知识 可以知道这个式子是 一个式子 的 展开式;
定义 res = 1;
res = res + res*a[i];
下面举例说明为什么
n = 3时
第一次循环:
res = 1 + 1*a[1];
第二次:
res = ( 1+a[1] )+ (1+a[1])*a[2] = 1+a[1] +a[2] + a[1]*a[2];
第三次:
res = (1+a[1]+a[2]+a[1]*a[2])+ ( 1+a[1]+a[2]+a[1]*a[2] ) * a[3] = 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];
其实 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];就是res = res + res*a[i]的展开式,一步步把括号去掉就是 这个展开式;
代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mod 1000000007
#define ll long long
int main()
{
ll i,j,n;
while(~scanf("%lld",&n))
{
ll ans = 1,t;
for(i = 0;i<n;i++)
{
scanf("%lld",&t);
ans = (ans + (ans*t)%mod)%mod;
}
printf("%lld\n",ans);
}
return 0;
}