1009. Product of Polynomials (25)

时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue

This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input 2 1 2.4 0 3.2 2 2 1.5 1 0.5 Sample Output 3 3 3.6 2 6.0 1 1.6

给出两个多项式的K个非零项的指数和系数，输出两式相乘后的非零项个数及各项的指数和系数。

系数结果保留一位小数。

#include<bits/stdc++.h> using namespace std; #define INF 0xfffffff #define MAXN 10010 int main() { #ifdef ONLINE_JUDGE #else freopen("F:/cb/read.txt","r",stdin); //freopen("F:/cb/out.txt","w",stdout); #endif ios::sync_with_stdio(false); cin.tie(0); int n; double a[MAXN],b[MAXN],ans[MAXN];//相乘数组开大 memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(ans,0,sizeof(ans)); cin>>n; int ma=0,mb=0,cnt=0; while(n--) { int x; double y; cin>>x>>y; ma=max(ma,x); a[x]=y; } cin>>n; while(n--) { int x; double y; cin>>x>>y; mb=max(mb,x); b[x]=y; } for(int i=0; i<=ma; ++i) for(int j=0; j<=mb; ++j) ans[i+j]+=(a[i]*b[j]); int pos=ma+mb; for(int i=ma+mb; i>=0; --i) if(ans[i]!=0) ++cnt,pos=min(pos,i); cout<<cnt<<" "; for(int i=ma+mb; i>pos; --i) if(ans[i]!=0) cout<<i<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<ans[i]<<" "; cout<<pos<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<ans[pos]<<endl; }