A. Two distinct points
You are given two segments [l1;r1] and [l2;r2] on the x-axis. It is guaranteed that l1<r1 and l2<r2. Segments may intersect, overlap or even coincide with each other.
Your problem is to find two integers a and b such that l1≤a≤r1, l2≤b≤r2 and a≠b. In other words, you have to choose two distinct integer points in such a way that the first point belongs to the segment [l1;r1] and the second one belongs to the segment [l2;r2].
It is guaranteed that the answer exists. If there are multiple answers, you can print any of them.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1≤q≤500) — the number of queries.
Each of the next q lines contains four integers l1i,r1i,l2i and r2i (1≤l1i,r1i,l2i,r2i≤109,l1i<r1i,l2i<r2i) — the ends of the segments in the i-th query.
Output
Print 2q integers. For the i-th query print two integers ai and bi — such numbers that l1i≤ai≤r1i, l2i≤bi≤r2i and ai≠bi. Queries are numbered in order of the input.
It is guaranteed that the answer exists. If there are multiple answers, you can print any.
Example
input
5
1 2 1 2
2 6 3 4
2 4 1 3
1 2 1 3
1 4 5 8
output
2 1
3 4
3 2
1 2
3 7
题意：给出L1, R1, L2, R2,求两个数满足L1 <= a <= R1, L2 <= b <= R2, a != b。
思路：先仔细读题。。开始我连a != b都没注意到真是@#%￥。为了节约时间考虑各种特判我直接粗暴的进行了两重for循环，，反正不会TLE。
代码：略
B. Divisors of Two Integers
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1,2,4,1,2,3,6]. Some of the possible lists are: [1,1,2,4,6,3,2], [4,6,1,1,2,3,2] or [1,6,3,2,4,1,2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2≤n≤128) — the number of divisors of x and y.
The second line of the input contains n integers d1,d2,…,dn (1≤di≤104), where di is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
input
10
10 2 8 1 2 4 1 20 4 5
output
20 8
题意：给出n个数，他们是两个数x, y的因子，求出x, y。特别地，如果某个数是x, y的公因子，它会出现两次。
思路：考虑到n不大，先粗暴地从大到小排序，根据性质易得当前最大数便是x, y其中较大一个；然后依次从数列中删去该数的因子，剩下的数中那个最大数即x, y其中较小一个。
代码：略
C. Nice Garland
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is si ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i,j such that ti=tj should be satisfied |i−j| mod 3=0. The value |x| means absolute value of x, the operation x mod y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
input
3
BRB
output
1
GRB
input
7
RGBGRBB
output
3
RGBRGBR
题意：包含RGB三个字符的一个数列，要求更换部分字符，使其任意相同字符之间的间距大于2，举个例子，RBGR，合法；RBR，不合法。
思路：很巧妙的一道题，注意到如果要满足RGB三个字符同时满足题目条件，只有诸如RGBRGBRG...或者GRBGRBGRB...的数列满足。这样我们可以暴力地打出6种有限循环数列的情况，然后和原数列进行比较，字符不同个数最少的数列为满足题意的答案。
代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define MAXN 200005 5 #define INF 1 << 30 6 7 const int num[6][3] = { 8 {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} 9 }; 10 11 int n, a[MAXN], tot, ans = INF, x; 12 char ch[MAXN]; 13 14 int main() { 15 scanf("%d", &n); 16 scanf("%s", ch + 1); 17 for (int i = 1; i <= n; i++) a[i] = ch[i] == 'R' ? 1 : ch[i] == 'G' ? 2 : 3; 18 for (int i = 0; i <= 5; i++) { 19 tot = 0; 20 for (int j = 1; j <= n; j++) tot += a[j] != num[i][j % 3]; 21 if (tot < ans) ans = tot, x = i; 22 } 23 printf("%d\n", ans); 24 for (int i = 1; i <= n; i++) printf(num[x][i % 3] == 1 ? "R" : num[x][i % 3] == 2 ? "G" : "B"); 25 }
D. Diverse Garland
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is si ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is diverse.
A garland is called diverse if any two adjacent (consecutive) lamps (i. e. such lamps that the distance between their positions is 1) have distinct colors.
In other words, if the obtained garland is t then for each i from 1 to n−1 the condition ti≠ti+1 should be satisfied.
Among all ways to recolor the initial garland to make it diverse you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a diverse garland from the given one.
In the second line of the output print one string t of length n — a diverse garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
input
9
RBGRRBRGG
output
2
RBGRGBRGR
input
8
BBBGBRRR
output
2
BRBGBRGR
input
13
BBRRRRGGGGGRR
output
6
BGRBRBGBGBGRG
题意：包含RGB三个字符的一个数列，要求更换部分字符，使其任意相同字符不相邻。
思路：D和C算是姊妹题吧，不过思路完全不一样，个人觉得D题思路更容易得出，主要是样例给得相当完美了。从样例中我们容易得出，我们可以对所有连着字符都相同的子串进行拆分，然后对其隔一个字符修改一个，对于连续个数为奇数的情况，修改非最左端右端的字符，修改后的字符只需满足非原字符即可，如GGGGG可修改为GBGBG；对于连续个数为偶数的情况，如果从最左端开始修改，则需考虑该连续字符子串左端的字符，修改后的字符应与其不同，如BGGGG可修改为BRGRG。不过可以不这样特殊考虑，详细见代码。
代码：
1 #include <bits/stdc++.h> 2 #define MAXN 200005 3 4 const char col[3] = {'R', 'G', 'B'}; 5 int n, ans, r; 6 char a[MAXN]; 7 8 int main() { 9 scanf("%d", &n); 10 scanf("%s", a + 1); 11 for (int l = 1; l < n;) { 12 if (a[l] == a[l + 1]) { 13 int tot = 0; 14 for (int j = l + 1; j <= n + 1; j++, tot++) 15 if (a[j] != a[l]) { 16 r = j - 1; 17 break; 18 } 19 for (int j = l - tot % 2 + 1; j <= r; j += 2) 20 for (int k = 0; k <= 2; k++) 21 if (col[k] != a[j] && col[k] != a[j - 1]) { 22 a[j] = col[k]; 23 break; 24 } 25 ans += (tot + 1)/ 2; 26 l = r + 1; 27 } 28 else l++; 29 } 30 printf("%d\n", ans); 31 for (int i = 1; i <= n; i++) printf("%c", a[i]); 32 return 0; 33 }