北京信息科技大学第十一届程序设计竞赛

Problem A kotori和糖果

https://ac.nowcoder.com/acm/contest/940/A

题意：合并，求代价最小值

题解：将一个堆二分，递归求该堆合并的最小代价，用map判重。

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;
ll t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
map<ll,ll>mp;
ll sloved(ll x){if(mp[x])return mp[x];if(x==0||x==1||( x & (x - 1)) == 0 )return mp[x]=0;if(x&1)return mp[x]=sloved(x>>1)+sloved((x>>1)+1)+1;elsereturn mp[x]=2*sloved(x>>1);
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%lld",&t);while(t--){scanf("%lld",&n);cout<<sloved(n)<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem B kotori和气球

https://ac.nowcoder.com/acm/contest/940/B

题意：求满足要求的n个气球的排列种数

题解：当仅当第一个可以选择n种颜色，其他不能与前面相同，所以只能选择m-1种颜色

$ans=n*(n-1)^{m-1}$

C++版本一

题解：快速幂+规律+数学

/*
*@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=109;
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 ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
ll powmod(ll a,ll b=MOD-2,ll c=MOD){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 DEBUGfreopen("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,&m);cout<<(n*powmod(n-1,m-1,MOD))%MOD<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

C++版本二

题解：dp递推，在第i个位置放颜色为x的球的方案数为在i-1处放除x以外的球的方案数之和
状态转移方程：s[i][x]=sum{s[i-1][y]，y!=x}

#include<bits/stdc++.h>
using namespace std;
#define PB push_back
#define LL long long
#define FI first
#define SE second
#define POP pop_back()
#define PII pair<LL,LL>
#define PCC pair<char,char>
#define endl '\n'
#define ls x<<1
#define rs x<<1|1
#define m(x) a[x].l+a[x].r>>1const int N=1e3+7;
const int INF=1e8,mod=109;LL a[N][N];
int n,m;int main()
{cin>>n>>m;for(int i=1;i<=n;i++){a[1][i]=1;}for(int i=2;i<=m;i++){for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){if(k==j)continue;a[i][j]=(a[i][j]+a[i-1][k])%mod;}}}LL ans=0;for(int i=1;i<=n;i++)ans+=a[m][i];ans%=mod;cout<<ans;return 0;
}

Problem C kotori和出道

https://ac.nowcoder.com/acm/contest/940/C

题意：约瑟夫环问题

题解：思维+规律+数学

C++版本一

题解：题目是经典的约瑟夫环问题

但是数据过大，不能用递归函数法解决

打表找规律可得，a是 $2^{a}\leqslant n$ 的最大值，则 $ans=2*(n-2^{a})+1$

/*
*@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;
ll t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
ll yuesefu(ll n,ll m=2){if(n == 1){return 0; //这里返回下标,从0开始，只有一个元素就是剩余的元素0}else{return (yuesefu(n-1,m) + m) % n; //我们传入的n是总共多少个数}
}
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%lld",&t);while(t--){scanf("%lld",&n);ans=0;m=n;while(m){ans++;m>>=1;}printf("%lld\n",(n-(1ll<<(ans-1)))*2+1);}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem D kotori和迷宫

https://ac.nowcoder.com/acm/contest/940/D

题意：问出口数量和最近出口的距离

题解：BFS

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 dir[4][2]={0,1,-1,0,1,0,0,-1};
int vis[50][50];
char str[50][50];
struct node{int x,y,step;
}x,f;
int main()
{
#ifdef DEBUGfreopen("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,&m);for(int i=1;i<=n;i++){scanf("%s",str[i]+1);for(int j=1;j<=m;j++){if(str[i][j]=='k')u=i,v=j;}}queue<node>q;x.x=u;x.y=v;x.step=0;ans=INF;vis[x.x][x.y]=1;q.push(x);while(!q.empty()){f=q.front();q.pop();//cout<<f.x<<" "<<f.y<<endl;if(str[f.x][f.y]=='e'){ans=min(ans,f.step);cnt++;continue;}for(int i=0;i<4;i++){x.x=f.x+dir[i][0];x.y=f.y+dir[i][1];x.step=f.step+1;if(1<=x.x&&x.x<=n&&1<=x.y&&x.y<=m&&str[x.x][x.y]!='*'&&!vis[x.x][x.y]){q.push(x);vis[x.x][x.y]=1;}}}if(ans==INF)cout<<-1<<endl;elsecout<<cnt<<" "<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem E kotori和素因子

https://ac.nowcoder.com/acm/contest/940/E

题意：

题解：线性筛出1-1000所有的质数，然后DFS求解最小值

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=1000+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 pre[M],prime[N],a[N],vis[N];
char str;
struct node{};
void dfs(int x,int now){if(x>n){ans=min(ans,now);return ;}for(int i=1;i<=cnt;i++){if(a[x]%pre[i]==0&&!vis[pre[i]]){vis[pre[i]]=1;dfs(x+1,now+pre[i]);vis[pre[i]]=0;}}
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);prime[0]=prime[1]=1;for(int i=2;i<M;i++){if(!prime[i]){pre[++cnt]=i;//cout<<i<<endl;}for(int j=1;j<=cnt&&i*pre[j]<M;j++){prime[i*pre[j]]=1;if(i%pre[j]==0)break;}}//cout<<cnt<<endl;//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);ans=INF;dfs(1,0);if(ans==INF)cout<<-1<<endl;elsecout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem F kotori和n皇后

https://ac.nowcoder.com/acm/contest/940/F

题意：

题解：规律+STL

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{ll x,y;
}e[N];
set<ll>sth,stl,st45,st135;
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&k);ans=INF;for(int i=1;i<=k;i++){scanf("%lld%lld",&e[i].x,&e[i].y);if(ans==INF&&(sth.count(e[i].x)||stl.count(e[i].y)||st45.count(e[i].x-e[i].y)||st135.count(e[i].x+e[i].y)))ans=i;sth.insert(e[i].x);stl.insert(e[i].y);st45.insert(e[i].x-e[i].y);st135.insert(e[i].x+e[i].y);}scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%d",&m);cout<<(ans<=m?"Yes":"No")<<endl;}
#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem G kotori和抽卡（二）

https://ac.nowcoder.com/acm/contest/940/G

题意：

题解：组合数学

C++版本一

题解： $ans=\frac{n!}{m!*(n-m)!}*\frac{4^{m}*1^{n-m}}{5^{n}}$

/*
*@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 DEBUGfreopen("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,&m);double ans=1.0;for(int i=m+1;i<=n;i++){ans*=0.2*i;}for(int i=1;i<=m;i++){ans*=0.8;}for(int i=1;i<=n-m;i++){ans/=i;}printf("%.4f\n",ans);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem H andy和购物

https://ac.nowcoder.com/acm/contest/940/H

题意：

题解：贪心

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;
double ans,cnt,flag,temp,sum;
ll a[N];
double b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("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++)scanf("%lld",&a[i]);for(int i=1;i<=n;i++)scanf("%lf",&b[i]);sort(a+1,a+n+1);sort(b+1,b+n+1);ans=0;for(int i=1;i<=n;i++)ans+=a[i]*b[n-i+1];printf("%.3f\n",ans);}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem I andy种树

https://ac.nowcoder.com/acm/contest/940/I

题意：

题解：差分

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,q,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("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,&q);for(int i=1;i<=q;i++)scanf("%d%d",&l,&r),a[l]++,a[r+1]--;for(int i=1;i<=n;i++)a[i]+=a[i-1],printf("%d%c",a[i]," \n"[i==n]);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem J andy的树被砍了

https://ac.nowcoder.com/acm/contest/940/J

题意：

题解：前缀和+二分

C++版本一

题解：对位置i，应该找i-n中cj前缀和超过hj的最小位置；

因此求出c的前缀和d；

二分查找 $d_{j-1}+h_{j}$ 的最小位置；

相当于所有树都种植在第一天，但是高度增加 $d_{j-1}$ ；

/*
*@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[N],cnt,flag,temp,sum;
ll d[N],h[N],c[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("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++)scanf("%lld",&h[i]);for(int i=1;i<=n;i++)scanf("%lld",&c[i]);for(int i=1;i<=n;i++)d[i]=c[i]+d[i-1];for(int i=1;i<=n;i++)printf("%lld%c",lower_bound(d+i,d+n+1,d[i-1]+h[i])-d," \n"[i==n]);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}