题目链接

传送门

简单数学题

题目

思路

前置知识：
\[ \sum_{i=m}^{n}C_{i}^{m}=C_{n+1}^{m+1} \]
此题化简:
\[ \begin{aligned} &\sum_{i=1}^{n}i\sum_{j=i}^{n}C_{j}^{i}& \\ =&\sum_{i=1}^{n}iC_{n+1}^{i+1}& \\ =&(n+1)\sum_{i=1}^{n}C_{n}^{i}-\sum_{i=1}^{n}C_{n+1}^{i+1}&\\ =&(n+1)(2^n-1)-(2^{n+1}-C_{n+1}^{0}-C_{n+1}^{1})&\\ =&(n-1)2^{n}+1 \end{aligned} \]

代码实现如下:

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> piL;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("in","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;LL n;LL qpow(LL x, LL n) {LL res = 1;while(n) {if(n & 1) res = res * x % mod;x = x * x % mod;n >>= 1;}return res;
}int main() {while(~scanf("%lld", &n)) {printf("%lld\n", ((n - 1) % mod * qpow(2, n) % mod + 1) % mod);}return 0;
}

Count

题目

思路

我们假设第n头牛的编号为\(f_{n}\)，由题意可知递推式为\(f_{n}=f_{n-1}+2 \times f_{n-2}+n^{3}\)，根据这个我们可以推得矩阵快速幂的系数矩阵为:
\[ \left\{ \begin{matrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 3 & 0 & 3 & 1 & 0 & 0 \\ 3 & 0 & 3 & 2 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 1 \end{matrix} \right\} \]

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> piL;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("in","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)const double eps = 1e-8;
const int mod = 123456789;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;int t;
LL n;
int f[10], a[10][10];void mulself(int a[10][10]) {int c[10][10];memset(c, 0, sizeof(c));for(int i = 0; i < 10; i++) {for(int j = 0; j < 10; j++) {for(int k = 0; k < 10; k++) {c[i][j] = (c[i][j] + (long long)a[i][k] * a[k][j]) % mod;}}}memcpy(a, c, sizeof(c));
}void mul(int f[10], int a[10][10]) {int c[10];memset(c, 0, sizeof(c));for(int i = 0; i < 10; i++) {for(int j = 0; j < 10; j++) {c[i] = (c[i] + (long long)f[j] * a[j][i] ) % mod;}}memcpy(f, c, sizeof(c));
}int main() {scanf("%d", &t);while(t--) {scanf("%lld", &n);f[0] = 2, f[1] = 1, f[2] = 8, f[3] = 4, f[4] = 2, f[5] = 1;a[0][0] = 1, a[0][1] = 1, a[0][2] = 0, a[0][3] = 0, a[0][4] = 0, a[0][5] = 0;a[1][0] = 2, a[1][1] = 0, a[1][2] = 0, a[1][3] = 0, a[1][4] = 0, a[1][5] = 0;a[2][0] = 1, a[2][1] = 0, a[2][2] = 1, a[2][3] = 0, a[2][4] = 0, a[2][5] = 0;a[3][0] = 3, a[3][1] = 0, a[3][2] = 3, a[3][3] = 1, a[3][4] = 0, a[3][5] = 0;a[4][0] = 3, a[4][1] = 0, a[4][2] = 3, a[4][3] = 2, a[4][4] = 1, a[4][5] = 0;a[5][0] = 1, a[5][1] = 0, a[5][2] = 1, a[5][3] = 1, a[5][4] = 1, a[5][5] = 1;if(n <= 2) {printf("%d\n", f[n]);continue;}n -= 2;while(n) {if(n & 1) mul(f, a);mulself(a);n >>= 1;}printf("%d\n", f[n]);}return 0;
}