A. Feibolaqi
Request $ f [n-1] * f [n + 1] -f [n] ^ 2 $, $ f [n] $ is a column number of Feibolaqi $ $ n-items
Count the first few difficult to find answers to the $ (- 1) ^ n $, prove it by mathematical induction the following:
$ N = 2 $, Conjecture
Suppose $ n = conjecture when k $, i.e. $ f [k-1] * f [k + 1] -f [k] ^ 2 = (- 1) ^ k $
When $ n = time $ K, $ F [K] F [K + 2] -f [K +. 1] ^ 2 = F [K]
Is proved
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
#include<cstring>
char s[1000005];
using namespace std;
int main(){
scanf("%s",s);
int l=strlen(s);
int z=s[l-1]-'0';
if(z%2==0)
cout<<1<<endl;
else cout<<-1<<endl;
}
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
using namespace std;
bool vis[10005];
int a[10005];
vector<int> ans;
int main(){
long long a,b;
cin>>a>>b;
cout<<a+b<<endl;
}
C. Sequence
Violence is like, first find each prefix at the sequence and, with map and indicates that the prefix exists for each query, first determine whether the inquired before k, inquired if no judgment, and then judge whether the sequence of k multiples, otherwise. 1 for $ (SUM / Z) ~ K (SUM / Z) $ and whether prefixes are present
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=100005;
int ans[maxn];
ll a[maxn];
map<ll,int> mp;
int main(){
int n,q;
scanf("%d%d",&n,&q);
ll sum=0;
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
sum+=a[i];
mp[sum]=1;
}
for(int i=1;i<=q;i++){
int z;
scanf("%d",&z);
if(z>n||sum%z!= 0 || years [z] == 2) {
printf("No\n");
continue;
}
if(ans[z]==1){
printf("Yes\n");
continue;
}
for(int i=1;i*(sum/z)<sum;i++){
if(mp[i*(sum/z)]!=1){
printf("No\n");
ans[z]=2;
break;
}
}
if(ans[z]!=2){
printf("Yes\n");
ans[z]=1;
}
}
}
Seeking lower diameter like
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=100005;
vector< pair<int,int> > g[maxn];
int d[maxn];
bool vis[maxn];
void dfs(int u){
vis[u]=1;
for(int i=0;i<g[u].size();i++){
int v=g[u][i].first;
if(!vis[v]){
d[v]=d[u]+g[u][i].second;
dfs(v);
}
}
}
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<n;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
g[x].push_back(make_pair(y,z));
g[y].push_back(make_pair(x,z));
}
// cout<<-1<<endl;
dfs(1);
int cnt;
long long dmax=0;
for(int i=1;i<=n;i++){
if(d[i]>dmax){
dmax=d[i];
cnt=i;
}
vis[i]=0;
d[i]=0;
}
dmax=0;
dfs(cnt);
for(int i=1;i<=n;i++){
if(d[i]>dmax){
dmax=d[i];
}
}
cout<<dmax*10+(1+dmax)*dmax/2<<endl;
}
E. Travel frog
Rise longest sequence, but feel that Italy has a problem, describe the topic should not rise longest sequence 233, n is relatively large, with $ O (n * log n) $ wording
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=100005;
int dp[maxn];
int a[maxn];
int n;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int cnt=0;
dp[0]=-1e9;
for(int i=1;i<=n;i++){
if(a[i]>=dp[cnt]){
dp[++cnt]=a[i];
}
else {
int z=upper_bound(dp+1,dp+1+cnt,a[i])-dp;
dp[z]=a[i];
}
}
cout<<cnt<<endl;
}
F. child hierarchy
Known meaning of the questions, answers sequence order-independent, so that a first row of the sequence order for the i-th sequence, is multiplied by the number of required C $ {-n-K. 1} ^ {}. 1--C {i-1} ^ {k -1} -C_ {ni} ^ {k-1} $. And 1e9 + 7 is a prime number, according to the Euler's formula $ a ^ {p-1} \ equiv1mod p $
You can get the answer
#include <cstdio>
#include <algorithm>
#include<vector>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const int maxn=1005;
ll pmod(ll a,ll b){
if(a==0) return 0;
if(b==0) return 1;
if(b==1) return a%mod;
ll ans=pmod(a,b/2);
ans=ans*ans%mod;
if(b&1)
return ans*a%mod;
return ans;
}
ll a[maxn];
ll c[maxn][maxn];
int main(){
for(int i=0;i<=1000;i++)
c[i][0]=1;
c[1][1]=1;
for(int i=1;i<=1000;i++)
c[i][i]=1;
for(int i=1;i<=1000;i++)
for(int j=1;j<i;j++)
c[i][j]=(c[i-1][j]+c[i-1][j-1])%(mod-1);
int t;
scanf("%d",&t);
while(t--){
int n,k;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+1+n);
ll ans=1;
for(int i=2;i<n;i++){
ll z=c[n-1][k-1];
if(n-i>=k-1)
z-=c[n-i][k-1];
if(i-1>=k-1)
z-=c[i-1][k-1];
z = ((z)% (mod-1) + mod-1)% (mod-1);
z=pmod(a[i],z);
ans=ans*z%mod;
}
printf("%lld\n",ans);
}
return 0;
}