2017 JUST Programming Contest 4.0

最新推荐文章于 2022-04-20 21:14:58 发布
最新推荐文章于 2022-04-20 21:14:58 发布
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本文链接：https://blog.youkuaiyun.com/sasuke__/article/details/78329766
本文精选了算法竞赛中的多个题目并提供了详细的解题思路及代码实现，涉及区间和计算、数列推导、字符串处理、矩阵操作等核心算法，旨在帮助读者理解并掌握算法竞赛中的常见问题解决技巧。

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A

题意：求所有区间的&和

思路：我们统计二进制中每一位的贡献，很显然如果连续一段区间二进制中某一位都为1，那么这个区间的任何子区间都满足答案，所以我们直接统计满足条件的子区间个数即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e6 + 10;
const LL INF = 1e9 + 10;
int num[qq];
int n;

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		scanf("%d", &n);
		for(int i = 1; i <= n; ++i) {
			scanf("%d", num + i);
		}
		LL ans = 0, f = -1;
		LL l, r, cnt;
		for(int i = 0; i < 23; ++i) {
			cnt = 0;
			f = -1;
			for(int j = 1; j <= n; ++j) {
				if(f == -1) {
					if(num[j] & (1 << i)) {
						l = j;
						f = 1;
					}
				} else {
					if(!((num[j] & (1 << i)))) {
						f = -1;
						r = j - 1;
						cnt += (r - l + 1) + (1 + r - l) * (r - l) / 2;
					}
				}
			}
			if(f != -1) {
				r = n;
				cnt += (r - l + 1) + (1 + r - l) * (r - l) / 2;
			}
			ans += cnt * (LL)(1 << i);
//			printf("%lld\n", cnt);
		}
		printf("%lld\n", ans);
	}
	return 0;
}


B

题意：ai = （a(i - 1) - i） % m，求整个数组a

思路：题意保证了至少一个不为-1，所以找个一个不为-1的，向前向后推一下即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 5e5 + 10;
const LL INF = 1e9 + 10;
int num[qq];
int n, m;

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		scanf("%d%d", &n, &m);
		int id = -1;
		for(int i = 1; i <= n; ++i) {
			scanf("%d", num + i);
			if(num[i] != -1)	id = i;
		}
		for(int i = id + 1; i <= n; ++i) {
			if(num[i] == -1)	num[i] = (num[i - 1] + 1) % m;
		}
		for(int i = id - 1; i >= 1; --i) {
			if(num[i] == -1) {
				if(num[i + 1] == 0)	num[i] = m - 1;
				else	num[i] = num[i + 1] - 1;
			}
		}
		for(int i = 1; i <= n; ++i) {
			printf("%d%c", num[i], i == n ? '\n' : ' ');
		}
	}		
	return 0;
}


C

队友写的

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const LL INF = 1e9 + 10;
struct Node{
	int x,p;
}a[qq],b[qq];
int c[qq];
bool cmp(const Node &u,const Node &v) {
	if(u.x==v.x) return u.p<v.p;
	return u.x<v.x;
}
int Bin(int key,int r) {
	int l=0,ans=0;
	while(l<=r) {
		int m=(l+r)>>1;
		if(b[m].x==key) {
			while(b[m+1].x==key) m++;
			return m;
		}
		else if(b[m].x<key) {
			ans=m;
			l=m+1;
		}
		else r=m-1;
	}
	return ans;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--) {
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++) {
			scanf("%d",&a[i].x);
			b[i].x=a[i].x;
			a[i].p=i;
			b[i].p=i;
		}
		sort(b,b+n,cmp);
		for(int i=0;i<n;i++) {
			int num1=MOD-a[i].x-1;
			int p1;
			if(b[0].x>num1) p1=-1;
			else p1=Bin(num1,n-1);
			int p2=n-1;
			if(b[p1].p==i) p1--;
			if(b[p2].p==i) p2--;
			if(p1<0) c[i]=(b[p2].x+a[i].x)%MOD;
			else c[i]=max((b[p1].x+a[i].x)%MOD,(b[p2].x+a[i].x)%MOD);
		}
		for(int i=0;i<n;i++) printf("%d%c",c[i],i==n-1?'\n':' ');
	}
	return 0;
}


D

题意：给出一个长度为n的字符串，然后q次询问，每次给你a b c，问你区间a b 内出现字符c的次数

思路：前缀维护一下讨论一下就行

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 5e5 + 10;
const LL INF = 1e9 + 10;
int num[qq][26], tot[26];
int n, m;
char st[qq];

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		scanf("%d%d", &n, &m);
		scanf("%s", st);
		int len = strlen(st);
		mst(num[0], 0);
		mst(tot, 0);
		for(int i = 0; i < n; ++i) {
			if(i != 0) {
				for(int j = 0; j < 26; ++j) {
					num[i][j] = num[i - 1][j];
				}
			}
			num[i][st[i] - 'a']++;
			tot[st[i] - 'a']++;
		}
		char s[5];
		while(m--) {
			int a, b;	scanf("%d%d", &a, &b);
			scanf("%s", s);
			a--, b--;
			LL ans = 0;
			if(b - a + 1 < n) {
				a %= n, b %= n;
				if(a <= b) {
						if(a - 1 < 0) {
							ans += num[b][s[0] - 'a'];
						} else {
							ans += num[b][s[0] - 'a'] - num[a - 1][s[0] - 'a'];
						}
					} else {
						if(a - 1 < 0) {
							ans += num[n - 1][s[0] - 'a'] + num[b][s[0] - 'a'];
						} else {
							ans += num[n - 1][s[0] - 'a'] - num[a - 1][s[0] - 'a'] + num[b][s[0] - 'a'];
						}
					}
			} else {
				LL tmp = (b - a + 1) / n;
				if((b - a + 1) % n == 0) {
					ans = tmp * tot[s[0] - 'a'];
				} else {
					ans = tmp * tot[s[0] - 'a'];
					a %= n, b %= n;
	//				printf("QQQ %lld\n", ans);
					if(a <= b) {
						if(a - 1 < 0) {
							ans += num[b][s[0] - 'a'];
						} else {
							ans += num[b][s[0] - 'a'] - num[a - 1][s[0] - 'a'];
						}
					} else {
						if(a - 1 < 0) {
							ans += num[n - 1][s[0] - 'a'] + num[b][s[0] - 'a'];
						} else {
							ans += num[n - 1][s[0] - 'a'] - num[a - 1][s[0] - 'a'] + num[b][s[0] - 'a'];
						}
					}
//					ans = (LL)tmp * tot[s[0] - 'a'] + (LL)(num[n - 1][s[0] - 'a'] - num[a - 1][s[0] - 'a']) + (num[b][s[0] - 'a']);
				}
			}
			printf("%lld\n", ans);
		}
	}

G

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e6 + 10;
const LL INF = 1e9 + 10;
int num[qq];
int pre[qq], suf[qq];

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		int n;	scanf("%d", &n);
		for(int i = 1; i <= n; ++i) {
			scanf("%d", num + i);
		}
		for(int i = 1; i <= n; ++i) {
			if(i == 1)	pre[i] = num[i];
			else	pre[i] = max(pre[i - 1], num[i]);
		}
		for(int i = n; i >= 1; --i) {
			if(i == n)	suf[i] = num[i];
			else	suf[i] = min(suf[i + 1], num[i]);
		}
		int cnt = 0;
		for(int i = 2; i < n; ++i) {
			if(pre[i - 1] <= num[i] && num[i] <= suf[i + 1]) {
				cnt++;
			}
		}
		printf("%d\n", cnt);
	}
	return 0;
}


H

题意：给出n*m的矩阵，问你能否将行数为1 n 列数为1 m的全部是1，每次可以交换两个位置的元素，求出最小交换次数，不可就输出-1

思路：很简单，判断一下1的个数即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 5e5 + 10;
const int INF = 1e9 + 10;
char st[105];

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		int n, m;	scanf("%d%d", &n, &m);
		int cnt1, cnt2, cnt3, cnt4;
		cnt1 = cnt2 = cnt3 = cnt4 = 0;
		for(int i = 1; i <= n; ++i) {
			scanf("%s", st + 1);
			for(int x, j = 1; j <= m; ++j) {
				x = st[j] - '0';
				if(i == 1 || j == 1 || i == n || j == m) {
					if(x == 0)	cnt1++;
				} else {
					if(x == 1)	cnt2++;
				}
			}
		}
		if(cnt2 >= cnt1) {
			printf("%d\n", cnt1);
		} else {
			puts("-1");
		}
	}
	return 0;
}



I

很简单的dp

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 5e5 + 10;
const LL INF = 1e9 + 10;
int num[qq], pos[qq];
int dp[qq];

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		int n;	scanf("%d", &n);
		for(int i = 1; i <= n; ++i) {
			scanf("%d", num + i);
		}
		mst(pos, -1);
		dp[1] = 0, pos[num[1]] = 1;
		for(int i = 2; i <= n; ++i) {
			dp[i] = dp[i - 1] + 1;
			if(pos[num[i]] != -1) {
				dp[i] = min(dp[i], dp[pos[num[i]]] + 1);
			}
			pos[num[i]] = i;
		}
		printf("%d\n", dp[n]);
	}	
	return 0;
}


J

队友写的

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const LL INF = 1e9 + 10;
LL a[qq];
int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++) scanf("%lld",&a[i]);
		if(n==1) {
			printf("%lld\n",a[0]%MOD);
			continue;
		}
		LL sum=(a[0]+a[1]+a[0]*a[1]%MOD)%MOD;
		for(int i=2;i<n;i++) {
			sum=(sum*(a[i]+1)%MOD+a[i])%MOD;
		}
		printf("%lld\n",sum%MOD);
	}		
	return 0;
}



K

队友写的

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const LL INF = 1e9 + 10;
LL a[30],F[15];
char s[25];
void init() {
	F[0]=1LL;
	for(int i=1;i<12;i++) F[i]=F[i-1]*1LL*i;
}
int main(){
	init();
	int t;	scanf("%d", &t);
	while(t--) {
		mst(a,0);
		int n;
		scanf("%d%s",&n,s);
		for(int i=0;i<n;i++) a[s[i]-'a']++;
		int cnt=0;
		for(int i=0;i<26;i++) {
			if(a[i]&1) cnt++;
			a[i]/=2LL;
		}
		if((n&1)&&cnt!=1||!(n&1)&&cnt!=0) {
			printf("0\n");
			continue;
		}
		LL sum=F[(n>>1)];
		for(int i=0;i<26;i++) sum/=F[a[i]];
		printf("%lld\n",sum);
	}		
	return 0;
}



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CCPC压力测试用了这套题，和一个队友一起搞了。初期签到还算比较顺利，还碰到了挺多有意思的题目，做的好爽，但还是犯了很多低级错误，对于杭电要时时刻刻关同步或者scanf，并且写代码还是要仔细些。 A. Matrix 这道题我懵逼，全靠队友搞出来，待补。 C. Vertex Deletion 树形DP 问无向图，删除若干点使得剩下图中没有孤立点的方案数。 对于一个点，有删除和未删除情况，未删除要考虑是否需要父节点。因而设计状态dp[i][0/1/2]， 分别表示点 i 删除、未删除且有子树、未删除且无子树。
2021 Shandong Provincial Collegiate Programming Contest 个人训练题解
m0_46527476的博客
06-10 1471
题目链接 目录 B Build Roads 题意 给定n(1≤n≤2⋅105)n(1\leq n\leq2\cdot10^5)n(1≤n≤2⋅105)个点的完全图，每个点有一个权值为ai(1≤ai≤2⋅105)a_i(1\leq a_i\leq2\cdot10^5)ai​(1≤ai​≤2⋅105)，两个点iii和jjj之间的边权为gcd(ai,aj)gcd(a_i,a_j)gcd(ai​,aj​)。题目给定点权aia_iai​随机生成的程序、范围以及随机种子，要求该图的最小生成树。 题解 首先如果点权范围L
Nafis' Programming Contest Judge Tool-开源
05-07
《Nafis' Programming Contest Judge Tool：开源的力量与无限可能》 编程竞赛是技术爱好者们展示才华、提升技能的重要舞台，而Nafis' Programming Contest Judge Tool正是一款专为此类活动设计的利器。这款工具以其...
PRO-Programming Contest
青烟绕指柔的博客
01-11 304
题目大意： n个人m个题目，每个题要r分钟完成。比赛有t分钟。给出每个人会做哪些题目，请你安排一个每个人在什么时候做什么题目，使得做出来的题目数最多。在做题数一样多的情况下，罚时尽量小。 应该很容易看出来是一个流的模型。每个人一共能做的题目数量是固定的。所以最大流即可。但是怎么让罚时最少呢？肯定是大家一起做题，而不是一个人做多道题。所以我们可以动态加边，每次都只加流量为1的边，直到最大流不增加为止...
Happy Programming Contest
Apollo的博客
07-30 285
In Zhejiang University Programming Contest, a team is called “couple team” if it consists of only two students loving each other. In the contest, the team will get a lovely balloon with unique color f...
2017 JUST Programming Contest 2.0 题解
weixin_34174422的博客
02-19 335
【题目链接】 A - On The Way to Lucky Plaza 首先，$n>m$或$k>m$或$k>n$就无解。 设$p = \frac{A}{B}$，$ans = C_{n - 1}^{k - 1}{\left( {\frac{A}{B}} \right)^{k}}{\left( {\frac{B-A}{B}} \right)^{n - k}} = ...
The 15th Jilin Provincial Collegiate Programming Contest
m0_62311106的博客
04-20 1266
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2019 JUST Programming Contest J. Grid Beauty
weixin_30567471的博客
04-20 254
J. Grid Beauty time limit per test 3.0 s memory limit per test 256 MB input standard input output standard output You are given a grid gg consisting of nn rows each of which is ...
6-4 Programming Contest (30分)
weixin_43301333的博客
01-15 4627
Bob will participate in a programming contest. There are altogether n problems in the contest. Unlike in PAT (Programming Ability Test), in a programming contest one can not obtain partial scores. For...
2018 AICCSA Programming Contest
PCCCCC的博客
08-09 479
A - Tree Game Gym - 101968A 思路：树的重心，当以树的重心为根节点时，其任一子树的节点数不大于 n / 2， 所以当只有一个重心时，谁先到达重心谁就能赢，所以我们以树的重心为根节点，求出每个节点的深度，只要先手玩家的深度 >= 后手玩家，那么先手玩家必赢，所以当只有一个重心时，所以ans = n * (n - 1) / 2, 因为当深度相同时， 玩家位...
AizuOJ 0282 Programming Contest（模拟）
weixin_44014982的博客
10-09 230
Programming Contest Time Limit : 1 sec, Memory Limit : 131072 KB Japanese プログラミングコンテスト 白虎大学では毎年プログラミングコンテストが開催されています。チームの総数をNとしたとき、チームには1からNのチームIDがそれぞれ割り当てられています。コンテストは全てのチームの得点が0の状態から開始し、L秒間連続して行われます...
NIKKEI Programming Contest 2019
FLY_WHITE的博客
01-29 793
比赛网址：https://atcoder.jp/contests/nikkei2019-qual A-Subscribers Problem Statement We conducted a survey on newspaper subscriptions. More specifically, we asked each of the NN respondents the following ...
2019 Sichuan Province Programming Contest
04-01
As an AI language model, I don't have the latest information on the 2019 Sichuan Province Programming Contest. However, I suggest checking official websites of the contest organizers, such as the Sichuan Provincial Department of Education or the Sichuan Province Computer Science Association, for more information. Additionally, contacting local universities or schools that may have participated in the contest in the past may also provide some insight.
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