洛谷 P1067 多项式输出
作者:
quiet_
,
2023-04-07 16:21:26
,
所有人可见
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阅读 156
//没有AC 只通过了1个数据但是形式貌似是对的
using namespace std;
const int N = 105;
int a[N];
int n;
int main()
{
cin >> n;
for(int i = 1 ; i <= n + 1 ; i ++) scanf("%d",&a[i]);
int flag = 1;
for(int i = 1 ; i <= n + 1 ; i ++)
{
if(flag == 1 && a[i] > 0 && i != n + 1)
{
printf("%dx^",a[i]);
printf("%d",n + 1 - i);
flag = 0;
}
else if(flag == 1 && a[i] < 0 && i != n + 1)
{
printf("-%dx^",-1*a[i]);
printf("%d",n + 1 - i);
flag = 0;
}
else if(flag == 0 && a[i] < 0 && i != n + 1)
{
printf("-");
printf("%dx^",-1*a[i]);
printf("%d",n + 1 - i);
}
else if(flag == 0 && a[i] > 0 && i != n + 1)
{
printf("+%dx^",a[i]);
printf("%d",n + 1 - i);
}
if(i == n + 1 && a[i] > 0) printf("+%d",a[i]),flag = 0;
else if(i == n + 1 && a[i] < 0) printf("-%d",-1 * a[i]),flag = 0;
}
if(flag == 1) printf("0");
return 0;
}