PAT1010 一元多项式求导 (25 分)

题目描述

C++解法

解法1

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
    int a[2000000]={0};
    int k,e,count=0;
    while(scanf("%d%d",&k,&e)!=EOF){
        a[e]=k;
    }
    a[0]=0;
    for(int i=1;i<=1000;i++){
        a[i-1]=a[i]*i;
        a[i]=0;
        if(a[i-1]!=0) count++;
    }
    if(count==0) printf("0 0");
    else{
        for(int i=1000;i>=0;i--){
            if(a[i]!=0){
                printf("%d %d",a[i],i);
                count--;
                if(count!=0) printf(" ");
            }
        }
    }
    return 0;
}

解法2

#include <iostream>
using namespace std;
int main() {
    int a, b, flag = 0;
    while (cin >> a >> b) {
        if (b != 0) {
            if (flag == 1) cout << " ";
            cout << a * b << " " << b - 1;
            flag = 1;
        }
    }
    if (flag == 0) cout << "0 0";
    return 0;
}

python解法

l = input().split()

o = []

for i in range(int(len(l)/2)):
    a_i = int(l[i*2])
    n_i = int(l[i*2+1])

    if a_i == 0:
        o.append('0 0')
    elif n_i == 0:
        continue
    elif n_i == 1:
        o.append(str(a_i))
        o.append('0')
    else:
        o.append(str(a_i*n_i))
        o.append(str(n_i-1))

if len(o) == 0:
    print('0 0')
else:
    print(' '.join(o))