【HNOI2013】数列

题面

题解

设({a_n})为差分数组，可以得到柿子：

[egin{aligned} ans &= sum_{a_1 = 1} ^ m sum_{a_2 = 1} ^ m cdots sum_{a_{k-1} = 1} ^ m (n - sum_{i = 1} ^ {k - 1} a_i) \ &= nm^{k - 1} - sum_{a_1 = 1} ^ m sum_{a_2 = 1} ^ m cdots sum_{a_{k - 1} = 1} ^ m sum_{i = 1} ^ {k - 1} a_i \ &= nm^{k - 1} - sum_{i = 1} ^ {k - 1} sum_{a_1 = 1} ^ m sum_{a_2 = 1} ^ m cdots sum_{a_{k - 1} = 1} ^ m a_i \ &= nm ^ {k - 1} - sum_{i = 1} ^ {k - 1} sum_{a_i = 1} ^ m a_i imes m ^ {k - 2} \ &= nm ^ {k - 1} - m^{k - 2}(k - 1) imes frac{m(m + 1)}2 end{aligned} ]

没了

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))

inline long long read()
{
	long long data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

long long N;
int n, k, m, p;
inline int Add(int x, int y) { return (x + y) % p; }
inline int Minus(int x, int y) { return (x - y + p) % p; }
inline int Mul(int x, int y) { return 1ll * x * y % p; }
inline int fastpow(int x, int y)
{
	int ans = 1;
	for(; y; y >>= 1, x = 1ll * x * x % p)
		if(y & 1) ans = 1ll * ans * x % p;
	return ans;
}

inline int S(int x) { return 1ll * x * (x + 1) / 2 % p; }
int main()
{
	N = read(), k = read(), m = read(), p = read();
	n = N % p, k %= p, m %= p;
	printf("%d
", Minus(Mul(n, fastpow(m, k - 1)),
				Mul(fastpow(m, k - 2), Mul(k - 1, S(m)))));
	return 0;
}