牛客小白月赛14
Problem A 简单计数
https://ac.nowcoder.com/acm/contest/879/A
题意：
题解：矩阵快速幂+构造矩阵
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100+10;
const int M=100000+10;
const int MOD=998244353;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int cnt,flag,temp,sum;
int a[N];
struct Matrix{
int n;
Matrix(int nn = 1):n(nn){ memset(a,0,sizeof(a));};
long long a[N][N];
void print(){
for(int i = 0;i <= n; ++i)
for(int j= 0;j <= n; ++j)
printf("%lld%c",a[i][j]," \n"[j==n]);
}
Matrix operator*(const Matrix &b)const{
Matrix c(n);
for(int i = 0;i <= n; ++i){
for(int j = 0;j <= n; ++j){
for(int k = 0;k <= n; ++k){
c.a[i][j] += a[i][k] * b.a[k][j];
c.a[i][j] %= MOD;
}
}
}
//c.print();
return c;
}
};
Matrix ans,fac;
ll POW(ll a,ll b,ll c){
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%c;
base=(base*base)%c;
b>>=1;
}
return res;
}
void MatrixPOW(ll k){
while(k){
if(k&1)ans=ans*fac;
fac=fac*fac;
k>>=1;
}
}
void init(){
ans.n = fac.n = 2;
ans.a[0][0] = 1;
ans.a[1][1]=1;
ans.a[0][1] = 0;
fac.a[0][0] = 0;
fac.a[0][1] = n-1;
fac.a[1][0] = 1;
fac.a[1][1] = n-2;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%lld%lld",&n,&k);
init();
//ans.print();
//fac.print();
MatrixPOW(k);
//ans.print();
cout<<ans.a[0][0]<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem B 投硬币
https://ac.nowcoder.com/acm/contest/879/B
题意：
题解：乘法逆元+概率
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const ll MOD=998244353;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
ll a[N];
char str;
struct node{};
ll POW(ll a,ll b,ll c){
//cout<<a<<" "<<b<<" "<<c<<endl;
if(a==0)
return 1;
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%c;
base=(base*base)%c;
b>>=1;
}
return res;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
a[1]=1;
a[0]=1;
for(int i=2;i<N;i++)
a[i]=(a[i-1]*i)%MOD;
scanf("%lld%lld%lld",&n,&k,&p);
ans=0;
for(int i=k;i<=n;i++){
ans=(ans+((a[n]*POW(p,i,MOD)%MOD)*POW((1-p+MOD)%MOD,n-i,MOD)%MOD)*POW(a[i]*a[n-i]%MOD,MOD-2,MOD)%MOD)%MOD;
}
cout<<ans<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem C 植树造林
https://ac.nowcoder.com/acm/contest/879/C
题意：
题解：数学+思维
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
cout<<(n%2?1:2)<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem D 签到题I
https://ac.nowcoder.com/acm/contest/879/D
题意：
题解：排序
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
cout<<a[k]<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem E 等比数列三角形
https://ac.nowcoder.com/acm/contest/879/E
题意：
题解：
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
ans = 0;
double e = (1 + sqrt(5)) / 2;
for(int i = 1; i * i <= n; i++)
for(int j = i; j <= e * i; j++)
if(__gcd(i, j) == 1) ans = (ans + n / j / j)%MOD;
cout << ans << endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define endl '\n'
const int mod=1e9+7;
LL n;
double e=(1.0+sqrt(5.0))/2;
LL b[5000];
int main()
{
while(cin>>n){
LL ans=n;
int sqr=sqrt(n);
for(int i=2;i<=sqr;i++){
b[i]=i-(int)(i/e)-1;
}
for(int i=2;i<=sqr;i++){
for(int j=i*2;j<=sqr;j+=i){
b[j]-=b[i];
}
ans=(ans+(n/(i*i))*b[i])%mod;
}
cout<<ans<<endl;
}
return 0;
} Problem F 乐色王传奇
https://ac.nowcoder.com/acm/contest/879/F
题意：
题解：
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=2500+10;
const int M=10000000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp=1,sum;
int a[N][N],inv[N];
int vis[N];
char str;
struct node{
int v,id;
node(){};
node(int _v,int _id){
v=_v,id=_id;
}
bool operator <(const node &S)const{
return v<S.v;
}
}e[M];
ll POW(ll a,ll b,ll c){
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%c;
base=(base*base)%c;
b>>=1;
}
return res;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&a[i][j]),e[(i-1)*n+j]=node(a[i][j],i);
sort(e+1,e+n*n+1);
int zercnt=n;inv[1]=1;
for(int i=2;i<=n;i++)
inv[i]=1LL*(MOD-MOD/i)*inv[MOD%i]%MOD;
for(int i=1;i<=n*n;i++){
if(!vis[e[i].id]++)
zercnt--;
if(vis[e[i].id]>1)
temp=1LL*temp*inv[vis[e[i].id]-1]%MOD*vis[e[i].id]%MOD;
if(!zercnt)
ans=(ans+1LL*temp*inv[vis[e[i].id]]%MOD*e[i].v)%MOD;
}
printf("%lld\n",1LL*ans*POW(POW(n,n,MOD),MOD-2,MOD)%MOD);
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem G many sum
https://ac.nowcoder.com/acm/contest/879/G
题意：
题解：朴素
C++版本一
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 5;
int a1[maxn];
long long b[maxn];
int main()
{
int n,a,m;
cin >> n >> a >> m;
a1[1] = a;
for(int i = 2; i <= n;i++){
a1[i] = (a1[i - 1] + 7 * i) % m;
}
for(int i = 1; i <= n;i++){
for(int j = i; j <= n; j+=i){
b[j]=b[j]+a1[i];
}
}
long long ans = 0;
for(int i = 1; i <= n;i++){
ans^=b[i];
}
cout << ans << endl;
}
Problem H 图上计数
https://ac.nowcoder.com/acm/contest/879/H
题意：
题解：
C++版本一
Problem I 有毒的玻璃球
https://ac.nowcoder.com/acm/contest/879/I
题意：
题解：积性函数+快速幂
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=10000000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
ll D[M];
int pre[M];
bool prime[M];
ll POW(ll a,ll b,ll c){
//cout<<a<<" "<<b<<" "<<c<<endl;
if(a==0)
return 1;
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%c;
base=(base*base)%c;
b>>=1;
}
return res;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d%d",&n,&k);
D[1]=1;
prime[0]=prime[1]=1;
for(int i=2;i<M;i++){
if(!prime[i]){
D[i]=POW(i,k,MOD);
pre[++cnt]=i;
}
for(int j=1;j<=cnt&&i*pre[j]<M;j++){
prime[i*pre[j]]=1;
D[i*pre[j]]=(D[i]*D[pre[j]])%MOD;
if(i%pre[j]==0){
break;
}
}
//cout<<i<<" "<<D[i]<<endl;
}
for(int i=1;i<=n;i++){
ans=(ans+(n/i)*D[i]%MOD)%MOD;
}
cout<<ans<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem J J.I
https://ac.nowcoder.com/acm/contest/879/J
题意：
题解：
C++版本一
