A.显然
代码语言：javascript复制#include <bits/stdc .h>
#define ll long long
using namespace std;
const int maxn = 200005;
int main()
{
int n;
cin>>n;
if(n%2==0){
n/=2;
cout<<(1<<n)<<endl;
}else {
cout<<0<<endl;
}
return 0;
}B.易得
代码语言：javascript复制//B
#include <bits/stdc .h>
#define ll long long
using namespace std;
char s[505][505];
int main()
{
int n,m,dian = 0;
cin>>n>>m;
for(int i=0;i<n;i ){
cin>>s[i];
for(int j=0;j<m;j ){
if(s[i][j]=='*')dian ;
}
}
int cent = 0,ni,nj,flag;
for(int i=0;i<n;i ){
for(int j=0;j<m;j ){
flag = 1;
if(s[i][j]=='*'); else flag = 0;
if(i-1>=0&&s[i-1][j]=='*'); else flag = 0;
if(j-1>=0&&s[i][j-1]=='*'); else flag = 0;
if(j 1<m&&s[i][j 1]=='*'); else flag = 0;
if(i 1<n&&s[i 1][j]=='*'); else flag = 0;
if(flag){
cent ;
ni = i;
nj = j;
}
}
}
if(cent!=1){
printf("NOn");
}else {
int tpi = ni,tpj = nj;
while(tpi>=0&&s[tpi][tpj]=='*'){
dian--; tpi--;
}
tpi = ni 1; tpj = nj;
while(tpi<n&&s[tpi][tpj]=='*'){
dian--; tpi ;
}
tpi = ni; tpj = nj - 1;
while(tpj>=0&&s[tpi][tpj]=='*'){
dian--; tpj--;
}
tpi = ni; tpj = nj 1;
while(tpj<m&&s[tpi][tpj]=='*'){
dian--; tpj ;
}
if(dian==0){
printf("YESn");
}else printf("NOn");
}
return 0;
}C.将数量相同但不是同一个结尾的放一个vector里，数量相同结尾相同放另一个vector里。
代码语言：javascript复制//C
#include <bits/stdc .h>
#define ll long long
using namespace std;
int n;
struct Node{
string s;
int num;
char last;
}node[100005];
int cmp(Node a,Node b){
if(a.num==b.num){
return a.last<b.last;
}else {
return a.num<b.num;
}
}
vector<pair<int,int> > pa1,pa2;
int main()
{
string s;
int num,size;
char last;
cin>>n;
for(int i=0;i<n;i ){
cin>>s; num = 0; size = s.size();
for(int i=0;i<size;i ){
if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u'){
num ; last = s[i];
}
}
node[i].num = num,node[i].last = last,node[i].s = s;
}
sort(node,node n,cmp);
int ls=-1;
int cnt = 1,pos = 0;
while(pos<n){
if(node[pos].num==node[pos 1].num&&node[pos].last==node[pos 1].last){
pa1.push_back(make_pair(pos,pos 1));
pos =2;
}else if(cnt!=node[pos].num){
ls = pos;
cnt = node[pos].num;
pos ;
}else if(ls==-1){
ls = pos;
cnt = node[pos].num;
pos ;
}else {
pa2.push_back(make_pair(pos,ls));
ls = -1;
pos ;
}
}
int ans,s1=pa1.size(),s2=pa2.size();
ans = min(s1,s2) max(0,(s1-s2)/2);
cout<<ans<<endl;
for(int i=0;i<min(s1,s2);i ){
cout<<node[pa2[i].first].s<<" "<<node[pa1[i].first].s<<endl;
cout<<node[pa2[i].second].s<<" "<<node[pa1[i].second].s<<endl;
}
if(s1>s2){
for(int i=min(s1,s2);i 1<s1;i =2){
cout<<node[pa1[i 1].first].s<<" "<<node[pa1[i].first].s<<endl;
cout<<node[pa1[i 1].second].s<<" "<<node[pa1[i].second].s<<endl;
}
}
return 0;
}E. 矩阵快速幂 欧拉降幂
第n项中f1的指数是fn-1 fn-2 fn-3之和，由此构造 。f2,f3同理
第n项中c的指数是 fn-1 fn-2 fn-3 2*i-6 ，构造矩阵
代码语言：javascript复制//E
#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
struct Mat{
ll m[101][101];
};
const ll mod = 1000000007;
int n;
Mat a;
Mat g,cx;
Mat mul(Mat x,Mat y,ll mo,int bound){
Mat cc;
for(int i=1;i<=bound;i ){
for(int j=1;j<=bound;j ){
cc.m[i][j] = 0;
for(int k=1;k<=bound;k ){
cc.m[i][j] = (cc.m[i][j] x.m[i][k]*y.m[k][j]%mo)%mo;
}
}
}
return cc;
}
Mat pow(Mat x,ll y,ll mo,int bound){
Mat ans;
for(int i=1;i<=bound;i )
for(int j=1;j<=bound;j )
if(i==j)ans.m[i][i] = 1;
else ans.m[i][j] = 0;
while(y){
if(y&1)ans = mul(ans,x,mo,bound);
x = mul(x,x,mo,bound);
y>>=1;
}
return ans;
}
ll quick(ll x,ll y){
ll re = 1;
while(y){
if(y&1)re = re*x%mod;
x=x*x%mod;
y>>=1;
}
return re;
}
int main(){
ll n,f1,f2,f3,c;
cin>>n>>f1>>f2>>f3>>c;
g.m[1][1] = 1; g.m[1][2] = 1; g.m[1][3] = 1;
g.m[2][1] = 1; g.m[2][2] = 0; g.m[2][3] = 0;
g.m[3][1] = 0; g.m[3][2] = 1; g.m[3][3] = 0;
cx.m[1][1] = 1; cx.m[1][2] = 1; cx.m[1][3] = 1; cx.m[1][4] = 1; cx.m[1][5] = 0;
cx.m[2][1] = 1; cx.m[2][2] = 0; cx.m[2][3] = 0; cx.m[2][4] = 0; cx.m[2][5] = 0;
cx.m[3][1] = 0; cx.m[3][2] = 1; cx.m[3][3] = 0; cx.m[3][4] = 0; cx.m[3][5] = 0;
cx.m[4][1] = 0; cx.m[4][2] = 0; cx.m[4][3] = 0; cx.m[4][4] = 1; cx.m[4][5] = 2;
cx.m[5][1] = 0; cx.m[5][2] = 0; cx.m[5][3] = 0; cx.m[5][4] = 0; cx.m[5][5] = 1;
ll exp_f1,exp_f2,exp_f3,num_c;
g = pow(g,n-3,1e9 6,3);
exp_f1 = g.m[1][3];
exp_f2 = g.m[1][2];
exp_f3 = g.m[1][1];
cx = pow(cx,n-3,1e9 6,5);
num_c = (cx.m[1][4]*2 cx.m[1][5])%(mod-1);
ll ans=0;
ans=quick(f1,exp_f1)%mod;
ans=ans*(quick(f2,exp_f2)%mod)%mod;
ans=ans*(quick(f3,exp_f3)%mod)%mod;
ans=ans*(quick(c,num_c)%mod)%mod;
cout<<ans<<endl;
return 0;
}
/*
5 1 2 5 3
*/
