P1067 多项式输出

一道有点复杂的模拟题

#include<cstdio>
using namespace std;
int a[102];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = n; i >= 0; i--)
        scanf("%d", &a[i]);
    for(int i = n; i >= 0; i--)
    {
        if(!a[i])
            continue;
        if(i > 1)
        {
            if(a[i] < 0 && a[i] != -1)
                printf("%dx^%d", a[i], i);
            else if(a[i] > 0 && a[i] != 1)
            {
                if(i != n)
                    printf("+");
                printf("%dx^%d", a[i], i);
            }
            else if(a[i] == 1)
            {
                if(i != n)
                    printf("+");
                printf("x^%d", i);
            }
            else if(a[i] == -1)
                printf("-x^%d", i);
        }
        else if(i == 1)
        {
            if(a[i] == 1 && i != n)
                printf("+x");
            else if(a[i] == -1 || i == n)
                printf("-x");
            else if((a[i] < 0 && a[i] != -1) || i == n)
                printf("%dx", a[i]);
            else if(a[i] > 0 && i != n && a[i] != 1)
                printf("+%dx", a[i]);
        }
        else if(!i)
        {
            if(a[i] < 0 || i == n)
                printf("%d", a[i]);
            else if(a[i] > 0 && i != n)
                printf("+%d", a[i]);
            else if(a[i] == 0 && n == 0)
                printf("0");
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/njbw7782/p/10356217.html