P1067 多项式输出-洛谷

解题思路:小模拟,认真考虑每种情况

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=105;
int n;
int a[N];
int main(){
  scanf("%d",&n);
  for(int i=0;i<=n;i++)scanf("%d",&a[i]);
  int i=0;
  int num=0;
  while(i<=n){
    if(a[i]!=0&&a[i]!=1&&a[i]!=-1){
        if(i!=0&&num){
         if(a[i]>0) printf("+");
        }
        printf("%d",a[i]);
        if(n-i!=0){
          if(n-i==1){
            printf("x");
          }
          else{
            printf("x^%d",n-i);
          }
        }
        num++;
    }
    else{
        if(a[i]==1||a[i]==-1){
            if(i!=0&&num&&a[i]>0){
                printf("+");
            }
            if(a[i]<0&&n-i!=0){
                printf("-");
            }
           if(n-i!=0){
             if(n-i==1){
                printf("x");
             }
             else{
               printf("x^%d",n-i);
             }
           }
           else{
            printf("%d",a[i]);
           }
          num++;
        }
    }
    i++;
  }
  if(num==0)
    printf("0");
return 0;
}

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转载自blog.csdn.net/lianghudream/article/details/84063185