Codeforces Round #162 (Div. 2) A~D 题解

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Codeforces Round #162 (Div. 2) A~D 题解

A. Colorful Stones (Simplified Edition)

time limit per test

memory limit per test

input

output

There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string s. The i-th (1-based) character of s represents the color of the i-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.

Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.

Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction c, if Liss is standing on a stone whose colors is c, Liss will move one stone forward, else she will not move.

You are given a string t. The number of instructions is equal to the length of t, and the i-th character of t represents the i-th instruction.

Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.

Input

The input contains two lines. The first line contains the string s (1 ≤ |s| ≤ 50). The second line contains the string t (1 ≤ |t| ≤ 50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.

Output

Print the final 1-based position of Liss in a single line.

Examples

Copy

RGB
RRR

Copy

RRRBGBRBBB
BBBRR

Copy

BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB
BBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB

Copy

15
模拟；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
string s, t;

int main()
{
	ios::sync_with_stdio(0);
	cin >> s >> t;
	int lens = s.length();
	int lent = t.length();
	int pos = 0;
	int st = 0;
	while (st < lent) {
		if (t[st] == s[pos]) {
			pos++; st++;
		}
		else st++;
	}
	printf("%d\n", pos + 1);
	return 0;
}

B. Roadside Trees (Simplified Edition)

time limit per test

memory limit per test

input

output

Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is h_i. Liss wants to eat all nuts.

Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:

Walk up or down one unit on a tree.
Eat a nut on the top of the current tree.
Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > h_i + 1.

Compute the minimal time (in seconds) required to eat all nuts.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁵) — the number of trees.

Next n lines contains the height of trees: i-th line contains an integer h_i (1 ≤ h_i ≤ 10⁴) — the height of the tree with the number i.

Output

Print a single integer — the minimal time required to eat all nuts in seconds.

Examples

Copy

2
1
2

Copy

Copy

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
int n;
int h[maxn];
int df[maxn];
int main()
{
//	ios::sync_with_stdio(0);
	n = rd();
	for (int i = 1; i <= n; i++) {
		h[i] = rd();
	}
	ll ans = 0;
	for (int i = 1; i < n; i++)df[i] = h[i + 1] - h[i];
	ans += 1ll*(h[1] + 1);
	for (int i = 2; i <= n; i++) {
		if (h[i] >= h[i - 1]) {
			ans += 1ll*(1 + 1 + h[i] - h[i - 1]);
		}
		else {
			ans += 1ll*(h[i - 1] - h[i] + 1 + 1);
		}
	}
	printf("%lld\n", ans);
	return 0;
}

C. Escape from Stones

time limit per test

memory limit per test

input

output

Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.

The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].

You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.

Input

The input consists of only one line. The only line contains the string s (1 ≤ |s| ≤ 10⁶). Each character in s will be either "l" or "r".

Output

Output n lines — on the i-th line you should print the i-th stone's number from the left.

Examples

Copy

llrlr

Copy

Copy

rrlll

Copy

Copy

lrlrr

Copy

Note

In the first example, the positions of stones 1, 2, 3, 4, 5 will be

Codeforces Round #162 (Div. 2) A~D 题解_#define

, respectively. So you should print the sequence: 3, 5, 4, 2, 1.

偏思维一点，找到规律就行了；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

string s;
int pos[maxn];
int main()
{
//	ios::sync_with_stdio(0);
	cin >> s;
	int lens = s.length();
	int ed = lens - 1;
	int st = 0;
	int tot = 0;
	for (int i = 0; i < lens; i++) {
		if (s[i] == 'l') {
			pos[ed] = (i + 1); ed--;
		}
		else {
			pos[st] = (i + 1); st++;
		}
	}
	for (int i = 0; i < lens; i++) {
		printf("%d\n", pos[i]);
	}
	return 0;
}

D. Good Sequences

time limit per test

memory limit per test

input

output

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a₁, a₂, ..., a_n are good.

Now she is interested in good sequences. A sequence x₁, x₂, ..., x_k is called good if it satisfies the following three conditions:

The sequence is strictly increasing, i.e. x_i < x_i + 1 for each i (1 ≤ i ≤ k - 1).
No two adjacent elements are coprime, i.e. gcd(x_i, x_i + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q).
All elements of the sequence are good integers.

Find the length of the longest good sequence.

Input

The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of good integers. The second line contains a single-space separated list of good integers a₁, a₂, ..., a_n in strictly increasing order (1 ≤ a_i ≤ 10⁵; a_i < a_i + 1).

Output

Print a single integer — the length of the longest good sequence.

Examples

Copy

5
2 3 4 6 9

Copy

9
1 2 3 5 6 7 8 9 10

Copy

Note

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

有趣的dp题目；

考虑枚举因数；

用dp[i]表示以因数i结尾时的最大值；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n;
int a[maxn];
int dp[maxn];
int ans;

void sol(int x) {
	int maxx = -inf;
	for (int i = 2; i <= sqrt(x); i++) {
		if (x%i == 0) {
			maxx = max(max(maxx, dp[i]), dp[x / i]);
		}
	}
	maxx = max(maxx, dp[x]);
	for (int i = 2; i <= sqrt(x); i++) {
		if (x%i == 0) {
			dp[i] = maxx + 1; dp[x / i] = maxx + 1;
			while (x%i == 0)x /= i;
		}
	}
	dp[x] = maxx + 1; ans = max(ans, maxx + 1);
}

int main()
{
//	ios::sync_with_stdio(0);
	n = rd();
	for (int i = 1; i <= n; i++) {
		a[i] = rd();
	}
	for (int i = 1; i <= n; i++)sol(a[i]);
	printf("%d\n", ans);
	return 0;
}