Codeforces Round #479 (Div. 3) 题解 977A 977B 977C 977D 977E 977F

A. Wrong Subtraction

题目大意:

\ 定义一种运算,让你去模拟

题解:

\ 模拟

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;

int n, k; 
int num[20], len;

int main()
{
	read(n), read(k);
	while(n) num[++ len] = n % 10, n /= 10;
	int i = 1;
	for(;k;-- k)
	{
		if(num[i]) -- num[i];
		else ++ i;
	}
	for(int j = len;j >= i;-- j) printf("%d", num[j]);
	return 0;
}

B. Two-gram

题目大意

\ 找在串S出现次数最多的串S的子串

题解

\ 暴力枚举,比较

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;

char s[1000], now;
int n, ans, cnt, p;

int main()
{
	read(n);
	scanf("%s", s + 1);
	for(now = 1;now < n;++ now)
	{
		cnt = 0;
		for(int j = 1;j < n;++ j)
			if(s[now] == s[j] && s[now + 1] == s[j + 1]) ++ cnt;
		if(cnt > ans) ans = cnt, p = now;
	}
	printf("%c%c", s[p], s[p + 1]);
	return 0;
}

C. Less or Equal

题目大意:

\ 给你一个数字序列和序列长度n,给你一个非负整数k,问是否存在一个数x,使得序列中小于等于x的元素的个数恰好为k。

题解:

\ 排序后看第k个和第k+1个是否相等。
\ 注意k=0的情况,需要特判。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;

int n, k, num[1000000];

int main()
{
	read(n), read(k);
	for(int i = 1;i <= n;++ i) read(num[i]);
	std::sort(num + 1, num + 1 + n);
	if(!k)
	{
		if(num[1] <= 1) printf("-1");
		else printf("1");
		return 0;
	}
	if(num[k] == num[k + 1]) printf("-1");
	else printf("%d", num[k]);
	return 0;
}

D. Divide by three, multiply by two

题目大意:

\ 给你一一堆数,让你把他们排序,要求对于相邻两个元素\(a_i,a_{i+1}\),满足\(a_I \times 2 = a_{i+1}\)\(a_i \div 3 = a_{i+1}\)保证答案存在

题解:

\ 由于2和3互质,所以对于任意一个数\(a\)\(a \div 3\)不可能通过\(\times 2\)\(\div 3\)操作变成\(a \times 2\)。这也就意味着如果把序列里每个数看做点,每个数向它能变成的序列里的数连一条边,一定是一条链。
\ 暴力建链,找头就行了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline long long max(long long a, long long b){return a > b ? a : b;}
inline long long min(long long a, long long b){return a < b ? a : b;}
inline long long abs(long long x){return x < 0 ? -x : x;}
inline void swap(long long &x, long long &y){long long tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f;

long long num[1000], nxt[1000], tag[1000], n;

int main()
{
	read(n);
	for(long long i = 1;i <= n;++ i) read(num[i]);
	for(long long i = 1;i <= n;++ i)
		for(long long j = 1;j <= n;++ j)
			if(i == j) continue;
			else if(num[i] * 2 == num[j] || num[i] == 3 * num[j]) nxt[i] = j, tag[j] = 1;
	for(long long i = 1;i <= n;++ i)
		if(!tag[i])
		{
			while(i) printf("%I64d ", num[i]), i = nxt[i];
			return 0;
		}
	return 0;
}

E. Cyclic Components

题目大意:

\ 给你一张图,问你里面有多个连通分量是一个简单环。

题解:

\ 按照找连通分量的过程dfs,如果一个点度数大于2,那么不可能形成简单环;如果每个点度数都小于2,且dfs过程中走回到某个已经走过的点了,就是简单环。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;

struct Edge
{
	int u, v, nxt;
	Edge(int _u, int _v, int _nxt){u = _u, v = _v, nxt = _nxt;}
	Edge(){}
}edge[2000010];
int head[2000010], cnt, vis[2000010], n, m, tmp1, tmp2, flag1, flag2, ans, fa[2000010];
inline void insert(int a, int b)
{
	edge[++ cnt] = Edge(a, b, head[a]), head[a] = cnt;
}
int find(int x)
{
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void dfs(int x, int pre)
{
	vis[x] = 1;
	int tot = 0;
	for(int pos = head[x];pos;pos = edge[pos].nxt)
	{
		++ tot;
		int v = edge[pos].v;
		if(v == pre) continue;
		int f1 = find(x), f2 = find(v);
		if(fa[f1] == fa[f2]) flag2 = 1;
		if(vis[v]) continue;
		else fa[f1] = f2;
		dfs(v, x);
	}
	if(tot > 2) flag1 = 0;
}

int main()
{
	read(n), read(m);
	for(int i = 1;i <= n;++ i) fa[i] = i;
	for(int i = 1;i <= m;++ i)
		read(tmp1), read(tmp2), insert(tmp1, tmp2), insert(tmp2, tmp1);
	for(int i = 1;i <= n;++ i)
		if(!vis[i])
		{
			flag1 = 1, flag2 = 0, dfs(i, -1);
			ans += flag1 & flag2;
		}
	printf("%d", ans);
	return 0;
}

F. Consecutive Subsequence

题目大意:

\ 给你一个序列,让你找一个单调递增的公差为1的最长子序列。

题解:

\ 先离散化,\(dp[i]\)表示以\(i\)为结尾的最长子序列长度,\(tong[i]\)表示数\(i\)在前面已经确定的\(dp\)状态中,数字为\(ti\)的位置上的最大的\(dp\)值。
\ 转移即可,用链表记录一下方案。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
	x = 0;char ch = getchar(), c = ch;
	while(ch < '0' || ch > '9') c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
	if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;

int dp[1000010], num[1000010], pos[1000010], pre[1000010], tong[1000010], n, tot, ans, p;
struct Node
{
	int rank, num;
}node[1000010];
bool cmp(Node a, Node b)
{
	return a.num < b.num;
}
void dfs(int x)
{
	if(!x) return;
	dfs(pre[x]);
	printf("%d ", x);
}
int main()
{
	read(n);
	for(int i = 1;i <= n;++ i) read(node[i].num), node[i].rank = i, dp[i] = 1;
	std::sort(node + 1, node + 1 + n, cmp);
	for(int i = 1;i <= n;++ i)
	{
		if(node[i].num != node[i - 1].num) ++ tot;
		if(node[i].num - node[i - 1].num > 1) ++ tot;
		pos[tot] = node[i].num, num[node[i].rank] = tot;
	}
	for(int i = 1;i <= n;++ i)
	{
		dp[i] += dp[tong[num[i] - 1]], pre[i] = tong[num[i] - 1];
		if(dp[tong[num[i]]] < dp[i])
			tong[num[i]] = i; 
	}
	for(int i = 1;i <= n;++ i)
		if(dp[i] > ans) ans = dp[i], p = i;
	printf("%d\n", ans);
	dfs(p);
	return 0;
}

针水。

posted @ 2018-05-08 08:31  嘒彼小星  阅读(473)  评论(0编辑  收藏