“美登杯”上海市高校大学生程序设计邀请赛 (华东理工大学)

Problem A 小花梨的字符串

https://acm.ecnu.edu.cn/contest/173/problem/A/

题意：对区间子字符串排列，使得满足条件，求排列最长。

题解： $\frac{(r-l+2)*(r-l+1)}{2}$

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[N];
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);cin>>str;for(int i=1;i<=m;i++){scanf("%d%d",&l,&r);cout<<(r-l+1)*(r-l+2)/2<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem B 小花梨的三角形

https://acm.ecnu.edu.cn/contest/173/problem/B/

题意：统计三个顶点不完全相同的三角形个数。

题解：枚举

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=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;
bool vis[N][N][N];
char str[N][N];
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+1;i++)scanf("%s",str[i]+1);for(int i=1;i<=n;i++){for(int j=1;j<=i;j++){for(int k=i+1;k<=n+1;k++){int a[4];a[0]=str[k][j]-'a';a[1]=str[i][j]-'a';a[2]=str[k][j+k-i]-'a';sort(a,a+3);//cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;vis[a[0]][a[1]][a[2]]=1;}}}for(int i=2;i<=n;i++){for(int j=1;j<i;j++){for(int k=j+1;k<=i&&i+k-j<=n+1;k++){int a[4];a[0]=str[i][j]-'a';a[1]=str[i][k]-'a';a[2]=str[i+k-j][k]-'a';sort(a,a+3);//cout<<i<<" "<<j<<" "<<k<<" "<<i+k-j<<endl;//cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;vis[a[0]][a[1]][a[2]]=1;}}}for(int i=0;i<26;i++){for(int j=i;j<26;j++){for(int k=j;k<26;k++){ans+=vis[i][j][k];}}}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem C 小花梨判连通

https://acm.ecnu.edu.cn/contest/173/problem/C/

题意：有K张联通图，求使得 ?和?在这 ?张图中都是连通的j的个数

C++版本一

题解：直接 dfs 、BFS 或者并查集对每张图进行连通块染色，或者并查集对每张图进行连通块染色，两个点在 k张图中都相邻说明这两个点在张图中都相邻说明这两个点在张图中都相邻说明这两个点在 k张图的染色都是一样的
map<vector<int>,int>存下每个点在 k张图的颜色序列出现的次数即可

/*
*@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=200000+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 pre[20][N];
char str;
bool vis[N];
struct node{};
vector<int>G[N];
vector<int>V[N];
map<vector<int>,int>mp;
int find(int *pre,int x){return pre[x]==x?x:pre[x]=find(pre,pre[x]);}
void dfs(int u){vis[u]=1;V[u].push_back(cnt);//cout<<u<<" "<<cnt<<endl;for(int i=0,j=G[u].size();i<j;i++){int v=G[u][i];if(!vis[v]){dfs(v);}}
}
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=0;i<=m;i++)for(int j=1;j<=n;j++)pre[i][j]=j;for(int i=1;i<=m;i++){scanf("%d",&k);memset(vis,0,sizeof(vis));for(int j=1;j<=n;j++)G[j].clear();cnt=0;for(int j=1;j<=k;j++){scanf("%d%d",&u,&v);int fx=find(pre[i],u);int fy=find(pre[i],v);if(fx!=fy){pre[i][fy]=fx;G[u].push_back(v);G[v].push_back(u);}}for(int j=1;j<=n;j++){if(vis[j]==0)++cnt,dfs(j);}}for(int i=1;i<=n;i++){mp[V[i]]++;}for(int i=1;i<=n;i++)cout<<mp[V[i]]<<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：对每一个图dfs求联通块并染色，处于同一个联通块中的点染成相同的颜色。染色之后对于图中的每一个点就有一个长度为k的颜色序列，两点在k张图中均联通即两个点的颜色序列完全相同。求每个点的颜色序列的hash值，并记录每种hash值出现了多少次。输出答案时将该点对应的hash值出现的次数输出即可。

#include "bits/stdc++.h"using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 100;
const int inf = 0x3f3f3f3f;
bool vis[maxn];
vector<int> e[11][maxn];
unsigned long long ans[maxn];
int c[11][maxn];
int col = 0;
map<unsigned long long, int> mp;void dfs(int deep, int now) {if (vis[now]) return;c[deep][now] = col;vis[now] = true;for (auto p:e[deep][now]) {dfs(deep, p);}
}int main() {//freopen("in.txt", "r", stdin);int n, k, m;scanf("%d %d", &n, &k);int u, v;for (int i = 0; i < k; i++) {cin >> m;for (int j = 0; j < m; j++) {scanf("%d %d", &u, &v);e[i][u].push_back(v);e[i][v].push_back(u);}col = 0;memset(vis, 0, sizeof(vis));for (int j = 1; j <= n; j++) {if (!vis[j]) col++;dfs(i, j);}}unsigned long long hash;for (int i = 1; i <= n; i++) {hash = 0;for (int j = 0; j < k; j++) {hash = hash * mod + c[j][i];}mp[hash]++;ans[i] = hash;}for (int i = 1; i <= n; i++) {cout << mp[ans[i]] << endl;}return 0;
}

Problem D 小花梨的取石子游戏

https://acm.ecnu.edu.cn/contest/173/problem/D/

题意：按规则取石子，求第i轮必胜的人

题解：博弈，谁先拿到非1的堆，谁就是赢家。因为足够聪明，所有人必然使得自己永远占先机。

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=200000+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],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("%d",&a[i]);a[i+n]=a[i];}flag=2*n+1;for(int i=2*n;i>=1;i--){if(a[i]!=1)flag=i;b[i]=flag;}for(int i=1;i<=n;i++){cout<<((min(n-1,b[i]-i))%2==0?"First":"Second")<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem E 小花梨的数组

https://acm.ecnu.edu.cn/contest/173/problem/E/

题意：

题解：线段树+tag标记

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=1000000+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 f[M],prime[M],pre[N];
ll tree[N<<2],tag[N<<2];
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 init(){f[1]=1;prime[0]=prime[1]=1;for(int i=2;i<=1e6;i++){if(prime[i]==0){pre[++cnt]=i;f[i]=i;}for(int j=1;j<=cnt&&pre[j]*i<=1e6;j++){prime[i*pre[j]]=1;f[i*pre[j]]=pre[j];if(i%pre[j]==0){break;}}}
}
void pushup(int rt){tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void pushdowm(int rt){tag[rt<<1]+=tag[rt];tag[rt<<1|1]+=tag[rt];tag[rt]=0;
}
void build(int l,int r,int rt){if(l==r){scanf("%lld",&tree[rt]);tag[rt]=0;return;}int mid=(l+r)>>1;build(l,mid,rt<<1);build(mid+1,r,rt<<1|1);pushup(rt);}
void updata1(int l,int r,int rt,int L,int R){if(tree[rt]==r-l+1)return;if(L<=l&&r<=R){tag[rt]++;return;}pushdowm(rt);int mid=(l+r)>>1;if(L<=mid)updata1(l,mid,rt<<1,L,R);if(R>mid)updata1(mid+1,r,rt<<1|1,L,R);pushup(rt);
}
void updata2(int l,int r,int rt,int L,int R){if(tree[rt]==r-l+1)return;if(L<=l&&r<=R&&tag[rt]){tag[rt]--;return;}if(l==r){tree[rt]=tree[rt]/f[tree[rt]]%MOD;return;}pushdowm(rt);int mid=(l+r)>>1;if(L<=mid)updata2(l,mid,rt<<1,L,R);if(R>mid)updata2(mid+1,r,rt<<1|1,L,R);pushup(rt);
}
ll query(int l,int r,int rt,int L){if(tree[rt]==r-l+1)return 1;if(l==r){return tree[rt]*POW(f[tree[rt]],tag[rt],MOD)%MOD;}pushdowm(rt);int mid=(l+r)>>1;if(L<=mid)return query(l,mid,rt<<1,L);else return query(mid+1,r,rt<<1|1,L);
}
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);init();while(~scanf("%d%d",&n,&m)){memset(tree,0,sizeof(tree));build(1,n,1);for(int i=1;i<=m;i++){scanf("%d",&k);if(k==1){scanf("%d%d",&l,&r);updata1(1,n,1,l,r);}else if(k==2){scanf("%d%d",&l,&r);updata2(1,n,1,l,r);}else{scanf("%d",&l);cout<<query(1,n,1,l)<<endl;}}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem F 小花梨的无向图

https://acm.ecnu.edu.cn/contest/173/problem/F/

题意：

题解：

C++版本一

Problem G 小花梨的函数

https://acm.ecnu.edu.cn/contest/173/problem/G/

题意：

题解：

C++版本一

Problem H 小花梨的矩阵

https://acm.ecnu.edu.cn/contest/173/problem/H/

题意：

题解：

C++版本一

Problem I 小花梨点外卖

https://acm.ecnu.edu.cn/contest/173/problem/I/

题意：有两种优惠，最多选择一种，使得付款最小

题解：模拟

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,a,b,c,d;
ll ans,cnt,flag,temp,sum;
int v[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%d%d%d",&n,&a,&b,&c,&d);for(int i=1;i<=n;i++){scanf("%d",&v[i]);sum+=v[i];}if(sum>=a&&sum>=c){ans=sum-max(b,d);}else if(sum>=a){ans=sum-b;}else if(sum>=c){ans=sum-d;}else{ans=sum;}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}