2019黑龙江大学程序设计竞赛

Problem A Find the Nth Character

https://ac.nowcoder.com/acm/contest/877/A

题意：定义一个字符串，求第n个字符是什么

题解：

打表

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;
char str[N];
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);
    for(int i=1;i<=200;i++){
        for(int j=1;j<=i;j++){
            str[++cnt]='a'+(j%26==0?26:j%26)-1;
        }
    }
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        cout<<str[n]<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem B Help Me

https://ac.nowcoder.com/acm/contest/877/B

题意：

题解：

完全平方公式，拆分，归类。

对于ai的平方各有n-1个

对于-2aiaj各1个，用前缀和可以减少复杂度

/*
*@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;
ll 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;
        sum=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            ans-=2*sum*a[i];
            sum+=a[i];
            ans+=(n-1)*a[i]*a[i];
        }
        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 Challenge IQ

https://ac.nowcoder.com/acm/contest/877/C

题意：在n的所有全排列数组中相邻互质对数最多有几对

题解：互质环

肯定存在一个n对的序列

就是1 2 3 4 5 6 7 。。。。。。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 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);
        printf("%d\n",n);
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem D Schedules

https://ac.nowcoder.com/acm/contest/877/D

题意：有n个任务，有固定的开始和结束时间，同一个机器不能同时进行两个任务，求至少需要几台机器

题解：

差分

原题

/*
*@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,u,v;
int ans,cnt,flag,temp,sum;
int l[N],r[N];
char str;
struct node{
    int l,r;
    bool operator <(const node &S)const{
        if(l==S.l)
            return r<S.r;
        return l<S.l;
    }
}e[N];
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--){
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++){
            //scanf("%d%d",&e[i].l,&e[i].r);
            scanf("%d%d",&l[i],&r[i]);
        }
        //sort(e+1,e+n+1);
        sort(l+1,l+n+1);
        sort(r+1,r+n+1);
        int x=1;
        int y=1;
        ans=0;
        t=0;
        for(int i=0;i<=N;i++){
            //while(e[x].l==i&&x<=n)x++,t++;
            //while(e[y].r==i&&y<=n)y++,t--;
            while(l[x]==i&&x<=n)x++,t++;
            while(r[y]==i&&y<=n)y++,t--;
            ans=max(ans,t);
        }
        cout<<ans<<endl;
    }

    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem E Pig-Palindromic

https://ac.nowcoder.com/acm/contest/877/E

题意：求一个长度为偶数，而且对称位置分别为大写小写字母或者小写大写字母的子字符串的长度

C++版本一

题解：

最长回文子串

改下条件==变成绝对值为32

/*
*@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  longestPalindrome(string s) {
	//寻找最长回文子串
	int sz = s.size();
	if (sz <= 1) return 0;
	//用动态规划方法
	//dp为size*size大小的矩阵，dp[i][j]表示以s[i]开头，s[j]结尾的回文串长度（如果不是回文串，则为0）
	vector<vector<int>> dp(sz);
	for (int i = 0; i<sz; i++) {
		for (int j = 0; j<sz; j++) {
			//初始化，将对角线元素设为1
			if(i==j) dp[i].push_back(1);
			else dp[i].push_back(0);
		}
	}
	int start = 0, maxl = 0;
	for (int j = 0; j < sz;j++){
		for (int i = j - 1; i >= 0; i--) {
			if (abs(s[i]-s[j])==32) {
				if(j-i==1) dp[i][j] = 2;
				else {
					if (dp[i + 1][j - 1]>0) {
						dp[i][j] = dp[i + 1][j - 1] + 2;
					}
					else dp[i][j] = 0;
				}
			}
			else dp[i][j] = 0;
			if (dp[i][j]%2==0&&dp[i][j]>maxl) {
				maxl = dp[i][j]; start = i;
			}
		}
	}
	return maxl;
}
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);
    cin>>str;
    cout<<longestPalindrome(str)<<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

题解：枚举相邻两点，向两边推；

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    string a;
    while(cin>>a)
    {
        int cnt;int maxn=-9999999;
    for(int i=0;i<a.size();i++)
    {
        cnt=0;
        for(int k=i,w=i+1;k>=0&&w<a.size();)
        {
            if(a[k]+32==a[w]||a[k]-32==a[w])
            {
                cnt+=2;k--;w++;
            }
            else
                break;
        } 
        maxn=max(maxn,cnt);
    }
    cout<<maxn<<endl;
    }
}

Problem F GCD Problem

https://ac.nowcoder.com/acm/contest/877/F

题解：一个数组，有0 1两种操作，0操作代表区间[l，r]中的所有元素变成，1操作代表求[l，r]的最大公约数

题解：线段树+结论

对于0操作每一个数最多操作8次，所以可以用线段树更新，只要标记一下，这个点已经不能更新了

/*
*@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;
char str;
struct node{};
ll tree[N<<2];
bool tag[N<<2];
void pushup(int rt){
    tree[rt]=__gcd(tree[rt<<1],tree[rt<<1|1]);
    tag[rt]=tag[rt<<1]&&tag[rt<<1|1];
}
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 updata(int l,int r,int rt,int L,int R){
    if(tag[rt])
        return;
    if(l==r){
        tree[rt]=sqrt(tree[rt]);
        if(tree[rt]==1)tag[rt]=1;
        return;
    }

    int mid=(l+r)>>1;
    if(L<=mid)updata(l,mid,rt<<1,L,R);
    if(R>mid)updata(mid+1,r,rt<<1|1,L,R);
    pushup(rt);
}
ll query(int l,int r,int rt,int L,int R){
    if(tag[rt]&&l<=L&&R<=r){
        return 1;
    }
    if(L<=l&&r<=R){
        return tree[rt];
    }
    int mid=(l+r)>>1;
    if(R<=mid)return query(l,mid,rt<<1,L,R);
    else if(L>mid)return query(mid+1,r,rt<<1|1,L,R);
    else return __gcd(query(l,mid,rt<<1,L,R),query(mid+1,r,rt<<1|1,L,R));
}
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);
    memset(tree,0,sizeof(tree));
    build(1,n,1);
    scanf("%d",&m);
    while(m--){
        scanf("%d%d%d",&t,&l,&r);
        if(t==0){
            updata(1,n,1,l,r);
        }else{
            cout<<query(1,n,1,l,r)<<endl;
        }
    }
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem G Bash Game

https://ac.nowcoder.com/acm/contest/877/G

题意：有一拍卖品成交价为P，两人轮流出价，至少加价1，最多加价M，两个人都足够聪明，求最后谁会拍下

题解：巴什博奕

当且仅当(n-1)%(m+1)==0时Bob能赢

参考文章：巴什博奕

/*
*@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,&m);
        cout<<((n-1)%(m+1)==0?"Bob":"Alice")<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem H Pass CET 6

https://ac.nowcoder.com/acm/contest/877/H

题意：每天只能背一行或者一列的单词，求一个代表单词的矩阵，其值为这个单词最后背是第几天

C++版本一

题解：

贪心

只标记行列最后一次，比较最后的行列最大值

/*
*@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;

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%d",&n,&m,&t);
    int a[n+10][m+10];
    for(int i=0;i<=n;i++)
        for(int j=0;j<=m;j++)
            a[i][j]=0;
    for(int i=1;i<=t;i++){
        scanf("%d%d",&k,&p);
        if(k==1){
            a[p][0]=i;
        }else{
            a[0][p]=i;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            a[i][j]=max(a[0][j],a[i][0]);
        }
        for(int j=1;j<=m;j++){
            printf("%d%c",a[i][j]," \n"[j==m]);
        }
    }
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem I Center Street

https://ac.nowcoder.com/acm/contest/877/I

题意：n个仓库，如果两个仓买A,B满足A是B的倍数，则A,B之间可以通一条路,当然这些路是双向的，A可以到B，B可以到A，如果仓买A,B之间通路，则过路费，从1号出发，求到每个仓库的最小花费，

题解：筛法

参考文章：线性筛

/*
*@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=500000+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;
ll dis[N];
int prime[N],tag[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--){
	memset(tag,0,sizeof(tag));
	int cnt=0;
	tag[0]=tag[1]=1;
	for(int i = 2; i<N; i++){
		if(!tag[i]){
            prime[cnt++]=i;
            dis[i]=(ll)(i-1)*(i-1);
		}
		for(int j=0;j<cnt && prime[j]*i<N; j++){
			tag[i*prime[j]] = 1;
			dis[i*prime[j]] = dis[i] + (ll)i*i*(prime[j]-1)*(prime[j]-1);
			if(i % prime[j]==0){
                break;
			}
		}
	}
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        printf("%lld%c",dis[i]," \n"[i==n]);
    }
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem J Special Distance

https://ac.nowcoder.com/acm/contest/877/J

题意：求最大FST距离

题解：

/*
*@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=300000+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;
ll a[N],b[N];
char str;
struct node{
ll x,y;
};
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;
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            a[i]*=a[i];
            b[i]=(ll)i*i;
        }
        node mx,mn;
        mx.x=1;mx.y=a[1];
        mn=mx;
        for(int i=2;i<=n;i++){
            ans=max(ans,a[i]-mn.y+b[i]-mn.x);
            ans=max(ans,mx.y-a[i]+b[i]-mx.x);
            if(a[i]-b[i]>mx.y-mx.x){
                mx.x=b[i];
                mx.y=a[i];
            }
            if(-a[i]-b[i]>-mn.y-mn.x){
                mn.x=b[i];
                mn.y=a[i];
            }
        }
        cout<<ans<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem K Maximum Sum of Digits

https://ac.nowcoder.com/acm/contest/877/K

题意：将n分成两个数，使得两个数每一位的数字加起来最大

题解：

CF原题

贪心

/*
*@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;
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("%lld",&n);
        if(n==0){
            cout<<0<<endl;
            continue;
        }
        m=n;
        ans=0;
        sum=0;
        while(n){
            sum++;
            n/=10;
        }
        ans=(sum-1)*9;
        m-=pow(10,sum-1)-1;
        while(m){
            ans+=m%10;
            m/=10;
        }
        cout<<ans<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}