【A】水题,输出 a + b + min(a,b),从来没看到过如此水的题。
【B】同样水题,统计一个只含有3个字符构成的字符串的价值,其中字符'H'的贡献为1,‘C’的贡献为12,'O'的贡献为16!
【C】矩阵快速幂,稍微推一下就可以得到下面矩阵了
【代码君】
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 2147493647;
struct Matrix{
LL a[8][8];
void init1(){
memset(a, 0, sizeof(a));
}
void init2(){
memset(a, 0, sizeof(a));
for(int i = 1; i <= 7; i++){
a[i][i] = 1;
}
}
};
Matrix operator*(const Matrix &a,const Matrix &b)
{
Matrix res;
res.init1();
for(int i = 1; i <= 7; i++){
for(int j = 1; j <= 7; j++){
for(int k = 1; k <= 7; k++){
res.a[i][j] = (res.a[i][j] + a.a[i][k]*b.a[k][j])%mod;
}
}
}
return res;
}
Matrix solve(Matrix a,LL n)
{
Matrix res;
res.init2();
while(n){
if(n%2) res = res*a;
a = a*a;
n>>=1;
}
return res;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
LL n,a,b;
scanf("%lld%lld%lld",&n,&a,&b);
Matrix A; A.init1();
Matrix B; B.init1();
Matrix C; C.init1();
A.a[1][1] = 1; A.a[1][2] = 2; A.a[1][3] = 1; A.a[1][4] = 1; A.a[1][5] = 4; A.a[1][6] = 6;A.a[1][7] = 4;
A.a[2][1] = 1; A.a[2][2] = 0; A.a[2][3] = 0; A.a[2][4] = 0; A.a[2][5] = 0; A.a[2][6] = 0;A.a[2][7] = 0;
A.a[3][1] = 0; A.a[3][2] = 0; A.a[3][3] = 1; A.a[3][4] = 0; A.a[3][5] = 0; A.a[3][6] = 0;A.a[3][7] = 0;
A.a[4][1] = 0; A.a[4][2] = 0; A.a[4][3] = 1; A.a[4][4] = 1; A.a[4][5] = 4; A.a[4][6] = 6;A.a[4][7] = 4;
A.a[5][1] = 0; A.a[5][2] = 0; A.a[5][3] = 1; A.a[5][4] = 0; A.a[5][5] = 1; A.a[5][6] = 3;A.a[5][7] = 3;
A.a[6][1] = 0; A.a[6][2] = 0; A.a[6][3] = 1; A.a[6][4] = 0; A.a[6][5] = 0; A.a[6][6] = 1;A.a[6][7] = 2;
A.a[7][1] = 0; A.a[7][2] = 0; A.a[7][3] = 1; A.a[7][4] = 0; A.a[7][5] = 0; A.a[7][6] = 0;A.a[7][7] = 1;
B.a[1][1] = b; B.a[2][1] = a; B.a[3][1] = 1; B.a[4][1] = 16; B.a[5][1] = 8; B.a[6][1] = 4; B.a[7][1] = 2;
//C = B*A^(n-2);
if(n==1){
printf("%lld\n",a%mod);
continue;
}
if(n==2){
printf("%lld\n",b%mod);
continue;
}
C=solve(A,n-2)*B;
printf("%lld\n",C.a[1][1]);
}
}
//
//Created by just_sort 2016/10/28
//Copyright (c) 2016 just_sort.All Rights Reserved
//
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define MP(x,y) make_pair(x,y)
const int maxn = 200005;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head
const int N = 105;
const int M = 1005;
struct edge{
int v,next;
}E[M];
bool vis[N],g[N][N];
int k, S;
LL ans;
int head[N],degree[N],point[N],edge_cnt;
void init(){
memset(head, -1, sizeof(head));
memset(g, false, sizeof(g));
memset(degree, 0 ,sizeof(degree));
edge_cnt = ans = 0;
}
void addedge(int u, int v){
E[edge_cnt].v = v, E[edge_cnt].next = head[u], head[u] = edge_cnt++;
}
void dfs(int u)
{
if(k == S)
{
ans++;
return ;
}
for(int i = head[u]; ~i; i = E[i].next)
{
int v = E[i].v;
int j;
if(degree[v] < S-1) continue;
for(j = 0; j < k; j++){
if(!g[v][point[j]]){
break;
}
}
if(j >= k)
{
point[k++] = v;
dfs(v);
point[--k] = 0;
}
}
}
int main()
{
int T, n, m;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d%d%d",&n,&m,&S);
for(int i = 1; i <= m; i++)
{
int u, v;
scanf("%d%d",&u,&v);
g[u][v] = g[v][u] = true;
degree[u]++;
degree[v]++;
if(u > v) swap(u, v);
addedge(u, v);
}
for(int i = 1; i <= n; i++){
if(degree[i] < S-1) continue;
k = 0;
point[k++] = i;
dfs(i);
}
printf("%lld\n",ans);
}
return 0;
}