2017 Multi-University Training Contest - Team 7：1010. Just do it（组合数？）

其他题目题解：

2017 Multi-University Training Contest - Team 7：1005. Euler theorem

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题意：有一个长度为n的整数序列，对其求m次前缀和，求最终的序列

令φ(x, y)为求完第x次前缀和后的第y个数，那么有

从这个公式可以得出两点：

①系数同于组合数

②系数为偶数的项值为0（异或性质a^a==0）

考虑对于组合数C(n,m)，当n&m==m(m<n)时为奇数

那么可以得知当n为2^n形式的整数时，C(n,m)当且仅当m=0和m=n时为奇数，其它都为偶数

这样就可以得出

因为是求第m个序列

所以对于φ(m, y)，对应着φ(m-2^k, y)，我们就可以用m&-m不停的取掉m的最低位，

然后将m&-m+1到y这一段的所有数字全部异或它前面的第m&-m个数字就好

~~一开始写的递归超时了~~

#include<stdio.h>
int a[200005], ans[200005];
int main(void)
{int T, n, m, k, i;scanf("%d", &T);while(T--){scanf("%d%d", &n, &m);for(i=1;i<=n;i++)scanf("%d", &a[i]);while(m){k = m&-m;for(i=k+1;i<=n;i++)a[i] ^= a[i-k];m -= k;}printf("%d", a[1]);for(i=2;i<=n;i++)printf(" %d", a[i]);printf("\n");}return 0;
}

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