PAT甲级-1002. A+B for Polynomials (25)多项式

素颜马尾好姑娘i 2022-05-31 09:19 144阅读 0赞

#  #

# 1002. A+B for 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**

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**

For each test case you should output the sum 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 to 1 decimal place.

**Sample Input**

2 1 2.4 0 3.2
    2 2 1.5 1 0.5

**Sample Output**

3 2 1.5 1 2.9 0 3.2

K是多项式非零项的个数，分别给出两个多项式的指数N和系数aN，计算两个多项式的值。

注意最后可能计算结果为全〇，要保留一位小数，零项不输出且输出式末尾不含空格。

#include<bits/stdc++.h>
    using namespace std;
    #define MAXN 1010
    #define INF 0xfffffff
    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);
        double a[MAXN],b[MAXN],ans[MAXN];
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(ans,0,sizeof(ans));
        int k,ma=-1,mb=-1,m=-1,mm=-1,cnt=0,mi=INF;
        cin>>k;
        while(k--)
        {
            int x;
            double y;
            cin>>x>>y;
            ma=max(ma,x);
            a[x]=y;
        }
        cin>>k;
        while(k--)
        {
            int x;
            double y;
            cin>>x>>y;
            mb=max(mb,x);
            b[x]=y;
        }
        m=max(ma,mb);
        for(int i=m; i>=0; --i)
        {
            if(a[i]==0&&b[i]==0) continue;
            if(a[i]+b[i]!=0)
            {
                ++cnt;
                mm=max(mm,i);
                mi=min(mi,i);
                ans[i]=a[i]+b[i];
            }
        }
        cout<<cnt;
        if(cnt>0)
        {
            cout<<" ";
            for(int i=m; i>mi; --i)
                if(ans[i]!=0)
                    cout<<i<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<ans[i]<<" ";
            cout<<mi<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<ans[mi]<<endl;
        }
        return 0;
    }