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[2013山东ACM省赛] Alice and Bob
Alice and Bob
Time Limit: 1000MS Memory limit: 65536K
题目描述
Alice and Bob like playing games very much.Today, they introduce a new game.
There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, Given an integer P, please tell the coefficient of the x^P.
Can you help Bob answer these questions?
输入
For each case, the first line contains a number n, then n numbers a0, a1, .... an-1 followed in the next line. In the third line is a number Q, and then following Q numbers P.
1 <= T <= 20
1 <= n <= 50
0 <= ai <= 100
Q <= 1000
0 <= P <= 1234567898765432
输出
示例输入
122 1234
示例输出
20
提示
来源
解题思路:
求多项式相乘展开式中x的某一指数的系数。
(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)给定这个式子,观察x的指数,2^0 2^1 2^2 。。很容易联想到二进制。
后来发现有规律,展开式中没有指数相同的两项,也就是说不能合并公因式。而某一x指数的系数化为二进制以后就可以找到规律了。
比如 求指数为13的系数,把13化为二进制 1 1 0 1 从右到左分别对应 a0 a1 a 2 a3 ,那么所求系数就是 a0 * a2 * a 3
再举例说明:
代码:
#include <iostream> #include <stack> #include <string.h> using namespace std; int a[51]; int t,n,q; long long p; int main() { cin>>t;while(t--) { cin>>n; for(int i=0;i<n;i++) cin>>a[i]; cin>>q; while(q--) { stack<int>s; cin>>p; int result=1; int cnt=-1; //这里使用了-1,为了方便,因为a数组是从0开始的 int yu; while(p) //把数化为二进制存到栈中 { cnt++; yu=p%2; s.push(yu); p/=2; } if(cnt>n-1) //当数的二进制位数大于n时,不存在直接输出0 { cout<<0<<endl; continue; } else { while(!s.empty()) { if(s.top()==1) { result*=a[cnt];//取数相乘 if(result>2012) result%=2012; } s.pop(); cnt--; } } cout<<result<<endl; } } return 0; }