题解 | 牛牛学数列
牛牛学数列
https://www.nowcoder.com/practice/0b97367cd2184c12a0e02f7c223aee11
#include <stdio.h>
int jisuan(int n)
{
int ret = 0;
for (int i = 1; i <= n; i++)
{
if (i % 2 == 0)
{
ret = ret - i;
}
else
{
ret = ret + i;
}
}
return ret;
}
int main()
{
int n;
scanf("%d", &n);
int ret = jisuan(n);
printf("%d", ret);
return 0;
}
理论上递归也能做，奈何我能力有限.......
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