AOJ/搜索与递归及分治法习题集

ALDS1_4_A-LinearSearch.

Description:

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Input:

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output:

Print C in a line.

Constraints:

n ≤ 10000

q ≤ 500

0 ≤ an element in S ≤ 109

0 ≤ an element in T ≤ 109

SampleInput1:

5

1 2 3 4 5

3

3 4 1

SampleOutput1:

3

SampleInput2:

3

3 1 2

1

5

SampleOutput2:

0

Codes:

//#define LOCAL

#include <cstdio>

int search(int A[], int n, int key) {
	int i = 0; A[n] = key;
	while(A[i] != key) ++i;
	return i!=n;
}

int main()
{
	#ifdef LOCAL
		freopen("E:\\Temp\\input.txt", "r", stdin);
		freopen("E:\\Temp\\output.txt", "w", stdout);
	#endif

	int i, n, q, key, sum = 0, A[10010];
	scanf("%d", &n);
	for(i=0; i<n; ++i) scanf("%d", &A[i]);
	scanf("%d", &q);
	for(i=0; i<q; ++i) {
		scanf("%d", &key);
		if(search(A, n, key)) ++sum;
	}
	printf("%d\n", sum);

	return 0;
}

ALDS1_4_B-BinarySearch.

Description:

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Input:

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output:

Print C in a line.

Constraints:

Elements in S is sorted in ascending order

n ≤ 100000

q ≤ 50000

0 ≤ an element in S ≤ 109

0 ≤ an element in T ≤ 109

SampleInput1:

5

1 2 3 4 5

3

3 4 1

SampleOutput1:

3

SampleInput2:

3

1 2 3

1

5

SampleOutput2:

0

Codes:

//#define LOCAL

#include <cstdio>

int n, A[1000010];

int binarySearch(int key) {
	int left = 0, right = n;
	while(left < right) {
		int mid = (left+right)/2;
		if(key > A[mid]) left = mid+1;
		else if(key == A[mid]) return 1;
		else right = mid;
	}
	return 0;
}

int main()
{
	#ifdef LOCAL
		freopen("E:\\Temp\\input.txt", "r", stdin);
		freopen("E:\\Temp\\output.txt", "w", stdout);
	#endif

	int i, q, key, sum = 0;
	scanf("%d", &n);
	for(i=0; i<n; ++i) scanf("%d", &A[i]);
	scanf("%d", &q);

	for(i=0; i<q; ++i) {
		scanf("%d", &key);
		if(binarySearch(key)) ++sum;
	}
	printf("%d\n", sum);

	return 0;
}

ALDS1_4_C-Dictionary.

Description:

Your task is to write a program of a simple dictionary which implements the following instructions:

insert str: insert a string str in to the dictionary

find str: if the distionary contains str, then print 'yes', otherwise print 'no'

Input:

In the first line n, the number of instructions is given. In the following n lines, n instructions are given in the above mentioned format.

Output:

Print yes or no for each find instruction in a line.

Constraints:

A string consists of 'A', 'C', 'G', or 'T'

1 ≤ length of a string ≤ 12

n ≤ 1000000

SampleInput1:

5

insert A

insert T

insert C

find G

find A

SampleOutput1:

no

yes

SampleInput2:

13

insert AAA

insert AAC

insert AGA

insert AGG

insert TTT

find AAA

find CCC

find CCC

insert CCC

find CCC

insert T

find TTT

find T

SampleOutput2:

yes

no

no

yes

yes

yes

Codes:

//#define LOCAL

#include <cstdio>
#include <cstring>

#define M 1046527
#define NIL (-1)
#define L 14
char H[M][L];

int getChar(char ch) {
	if(ch == 'A') return 1;
	else if(ch == 'C') return 2;
	else if(ch == 'G') return 3;
	else if(ch == 'T') return 4;
	else return 0;
}

long long getKey(char str[]) {
	int len = strlen(str);
	long long sum = 0, p = 1, i;
	for(i=0; i<len; ++i) {
		sum += p*(getChar(str[i]));
		p *= 5;
	}
	return sum;
}

int h1(int key) {return key%M;}
int h2(int key) {return 1+(key%(M-1));}

int find(char str[]) {
	long long key, i, h;
	key = getKey(str);
	for(i=0; ; ++i) {
		h = (h1(key)+i*h2(key))%M;
		if(strcmp(H[h], str) == 0) return 1;
		else if(strlen(H[h]) == 0) return 0;
	}
	return 0;
}

int insert(char str[]) {
	long long key, i, h;
	key = getKey(str);
	for(i=0; ; ++i) {
		h = (h1(key)+i*h2(key))%M;
		if(strcmp(H[h], str) == 0) return 1;
		else if(strlen(H[h]) == 0) {
			strcpy(H[h], str);
			return 0;
		}
	}
	return 0;
}

int main()
{
	#ifdef LOCAL
		freopen("E:\\Temp\\input.txt", "r", stdin);
		freopen("E:\\Temp\\output.txt", "w", stdout);
	#endif

	int i, n, h;
	char str[L], com[9];
	for(i=0; i<M; ++i) H[i][0] = '\0';
	scanf("%d", &n);

	for(i=0; i<n; ++i) {
		scanf("%s %s", com, str);
		if(com[0] == 'i') insert(str);
		else {
			if(find(str)) printf("yes\n");
			else printf("no\n");
		}
	}

	return 0;
}

ALDS1_4_D-Allocation.

Codes:

#include <iostream>
using namespace std;

#define MAX 100000
typedef long long llong;
int n, k; llong T[MAX];

int check(llong P) {
	int i = 0;
	for(int j=0; j<k; ++j) {
		llong s = 0;
		while(s+T[i] <= P) {
			s += T[i++];
			if(i == n) return n;
		}
	}
	return i;
}

int solve() {
	llong mid, left = 0, right = 1000000000;
	while(right-left > 1) {
		mid = (left+right)/2;
		int v = check(mid);
		if(v >= n) right = mid;
		else left = mid;
	}
	return right;
}

int main()
{
	cin >> n >> k;
	for(int i=0; i<n; ++i) cin >> T[i];
	cout << solve() << endl;
}

ALDS1_5_A-ExhaustiveSearch.

Description:

Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input:

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output:

For each question Mi, print yes or no.

Constraints:

n ≤ 20

q ≤ 200

1 ≤ elements in A ≤ 2000

1 ≤ Mi ≤ 2000

SampleInput:

5

1 5 7 10 21

8

2 4 17 8 22 21 100 35

SampleOutput:

no

no

yes

yes

yes

yes

no

no

Codes:

//#define LOCAL

#include <cstdio>

int n, A[50];

int solve(int i, int k) {
	if(!k) return 1;
	if(i >= n) return 0;
	return solve(i+1, k)||solve(i+1, k-A[i]);
}

int main()
{
	#ifdef LOCAL
		freopen("E:\\Temp\\input.txt", "r", stdin);
		freopen("E:\\Temp\\output.txt", "w", stdout);
	#endif

	int i, q, k;
	scanf("%d", &n);
	for(i=0; i<n; ++i) scanf("%d", &A[i]);
	scanf("%d", &q);

	for(i=0; i<q; ++i) {
		scanf("%d", &k);
		if(solve(0, k)) printf("yes\n");
		else printf("no\n");
	}

	return 0;
}

ALDS1_5_C-KochCurve.

Description:

Write a program which reads an integer n and draws a Koch curve based on recursive calles of depth n.

The Koch curve is well known as a kind of fractals.

You can draw a Koch curve in the following algorithm:

Divide a given segment (p1, p2) into three equal segments.

Replace the middle segment by the two sides of an equilateral triangle (s, u, t) of the same length as the segment.

Repeat this procedure recursively for new segments (p1, s), (s, u), (u, t), (t, p2).

You should start (0, 0), (100, 0) as the first segment.

Input:

An integer n is given.

Output:

Print each point (x, y) of the Koch curve. Print a point in a line. You should start the point(0, 0), which is the endpoint of the first segment and end with the point (100, 0), the other endpoint so that you can draw the Koch curve as an unbroken line. Each solution should be given as a decimal with an arbitrary number of fractional digits, and with an absolute error of at most 10-4.

Constraints:

0 ≤ n ≤ 6

SampleInput1:

1

SampleOutput1:

0.00000000 0.00000000

33.33333333 0.00000000

50.00000000 28.86751346

66.66666667 0.00000000

100.00000000 0.00000000

SampleInput2:

2

SampleOutput2:

0.00000000 0.00000000

11.11111111 0.00000000

16.66666667 9.62250449

22.22222222 0.00000000

33.33333333 0.00000000

38.88888889 9.62250449

33.33333333 19.24500897

44.44444444 19.24500897

50.00000000 28.86751346

55.55555556 19.24500897

66.66666667 19.24500897

61.11111111 9.62250449

66.66666667 0.00000000

77.77777778 0.00000000

83.33333333 9.62250449

88.88888889 0.00000000

100.00000000 0.00000000

Codes:

//#define LOCAL

#include <cstdio>
#include <cmath>

struct Point{ double x, y;};

void koch(int n, Point a, Point b) {
	if(!n) return ;

	Point s, t, u;
	double th = M_PI*60.0/180.0;

	s.x = (2.0*a.x+1.0*b.x)/3.0;
	s.y = (2.0*a.y+1.0*b.y)/3.0;
	t.x = (1.0*a.x+2.0*b.x)/3.0;
	t.y = (1.0*a.y+2.0*b.y)/3.0;
	u.x = (t.x-s.x)*cos(th)-(t.y-s.y)*sin(th)+s.x;
	u.y = (t.x-s.x)*sin(th)+(t.y-s.y)*cos(th)+s.y;

	koch(n-1, a, s);
	printf("%.8f %.8f\n", s.x, s.y);
	koch(n-1, s, u);
	printf("%.8f %.8f\n", u.x, u.y);
	koch(n-1, u, t);
	printf("%.8f %.8f\n", t.x, t.y);
	koch(n-1, t, b);
}

int main()
{
	#ifdef LOCAL
		freopen("E:\\Temp\\input.txt", "r", stdin);
		freopen("E:\\Temp\\output.txt", "w", stdout);
	#endif

	int n; Point a, b;
	scanf("%d", &n);
	a.x = 0, a.y = 0, b.x = 100, b.y = 0;

	printf("%.8f %.8f\n", a.x, a.y);
	koch(n, a, b);
	printf("%.8f %.8f\n", b.x, b.y);

	return 0;
}