PAT甲级-1009. Product of Polynomials (25)一元多项式相乘

绝地灬酷狼 2022-05-31 09:45 146阅读 0赞

# 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 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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;
    }