100pts求助（WA#1）

P1067 [NOIP2009 普及组] 多项式输出

V_HAPPY666 @ 2024-08-22 14:46:05

#include <bits/stdc++.h>
using namespace std;

int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int n;
    cin>>n;
    int xishu[n+1],k=n;
    for(int i=0;i<=n;i++)cin>>xishu[i];
    //最高次数项
    if(abs(xishu[0])==1)
    {
        if(xishu[0]==-1)cout<<'-';
    }
    else cout<<xishu[0];
    cout<<"x^"<<n--;
    //中间部分
    for(int i=1;i<k;i++)
    {
        int x=xishu[i];
        if(x>0)
        {
            cout<<'+';
            if(x==1)
            {
                if(n>1)cout<<"x^"<<n;
                else cout<<'x';
                //continue;
            }
            else
            {
                if(n>1)cout<<x<<"x^"<<n;
                else cout<<x<<'x';
                //continue;
            }
        }
        else if(x<0)
        {
            if(x==-1)
            {
                if(n>1)cout<<"-x^"<<n;
                else cout<<"-x";
                //continue;
            }
            else
            {
                if(n>1)cout<<x<<"x^"<<n;
                else cout<<x<<'x';
                //continue;
            }
        }
        n--;
    }
    //系数
    if(xishu[k]>0)cout<<'+';
    if(xishu[k]!=0)cout<<xishu[k];
    return 0;
}

by V_HAPPY666 @ 2024-08-22 14:46:30

谁能告诉我测试点1的数据是什么？

by 违规用户名K&xs3Z^ @ 2024-08-22 14:48:13

@V_HAPPY666 宝贵的一次下载。。求关


in:
0
1
out：
1

by V_HAPPY666 @ 2024-08-22 14:51:20

@违规用户名K&xs3Z^ 谢谢，关注了

by V_HAPPY666 @ 2024-08-22 14:51:43

@违规用户名K&xs3Z^ 没有看到下面n可以为0

by 违规用户名K&xs3Z^ @ 2024-08-22 14:52:49

@V_HAPPY666 嗯嗯 刚给你改到一半（悲

by V_HAPPY666 @ 2024-08-22 14:54:22

@违规用户名K&xs3Z^ 谢谢已经AC了

by 违规用户名K&xs3Z^ @ 2024-08-22 14:54:47

@V_HAPPY666 ok

by XIAOHAOYU1 @ 2024-09-18 18:44:44

?

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

似乎不需要那么长 求关

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