aaa

src/codeforces/set0/set1/set100/set160/set166/c/problem.md

@@ -0,0 +1,22 @@
+A median in an array with the length of n is an element which occupies position number after we sort the elements in the
+non-decreasing order (the array elements are numbered starting with 1). A median of an array (2, 6, 1, 2, 3) is the
+number 2, and a median of array (0, 96, 17, 23) — the number 17.
+
+We define an expression as the integer part of dividing number a by number b.
+
+One day Vasya showed Petya an array consisting of n integers and suggested finding the array's median. Petya didn't even
+look at the array and said that it equals x. Petya is a very honest boy, so he decided to add several numbers to the
+given array so that the median of the resulting array would be equal to x.
+
+Petya can add any integers from 1 to 105 to the array, including the same numbers. Of course, he can add nothing to the
+array. If a number is added multiple times, then we should consider it the number of times it occurs. It is not allowed
+to delete of change initial numbers of the array.
+
+While Petya is busy distracting Vasya, your task is to find the minimum number of elements he will need.
+
+### ideas
+
+1. sort it of course
+2. 至多增加n个x
+3. 需要知道有多少个a = 大于x, b = 小于x, c= 等于x的部分
+4. 让x为中位数，需要 b + c >= (a + b + c) / 2, (a + c) >= (a + b + c) / 2

src/codeforces/set0/set1/set100/set160/set166/c/solution.go

@@ -0,0 +1,106 @@
+package main
+
+import (
+	"bufio"
+	"fmt"
+	"os"
+)
+
+func main() {
+	reader := bufio.NewReader(os.Stdin)
+	n, x := readTwoNums(reader)
+	a := readNNums(reader, n)
+	res := solve(a, x)
+
+	fmt.Println(res)
+}
+
+func readInt(bytes []byte, from int, val *int) int {
+	i := from
+	sign := 1
+	if bytes[i] == '-' {
+		sign = -1
+		i++
+	}
+	tmp := 0
+	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
+		tmp = tmp*10 + int(bytes[i]-'0')
+		i++
+	}
+	*val = tmp * sign
+	return i
+}
+
+func readString(reader *bufio.Reader) string {
+	s, _ := reader.ReadString('\n')
+	for i := 0; i < len(s); i++ {
+		if s[i] == '\n' || s[i] == '\r' {
+			return s[:i]
+		}
+	}
+	return s
+}
+
+func readNum(reader *bufio.Reader) (a int) {
+	bs, _ := reader.ReadBytes('\n')
+	readInt(bs, 0, &a)
+	return
+}
+
+func readTwoNums(reader *bufio.Reader) (a int, b int) {
+	res := readNNums(reader, 2)
+	a, b = res[0], res[1]
+	return
+}
+
+func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
+	res := readNNums(reader, 3)
+	a, b, c = res[0], res[1], res[2]
+	return
+}
+
+func readNNums(reader *bufio.Reader, n int) []int {
+	res := make([]int, n)
+	x := 0
+	bs, _ := reader.ReadBytes('\n')
+	for i := 0; i < n; i++ {
+		for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
+			x++
+		}
+		x = readInt(bs, x, &res[i])
+	}
+	return res
+}
+
+func solve(a []int, x int) int {
+	cnt := make([]int, 3)
+
+	for _, num := range a {
+		if num < x {
+			cnt[0]++
+		} else if num == x {
+			cnt[1]++
+		} else {
+			cnt[2]++
+		}
+	}
+	cnt[0] += cnt[1]
+	cnt[2] += cnt[1]
+	n := len(a)
+	for add := 0; add <= n; add++ {
+		// add x
+		if (add+n)%2 == 1 {
+			// 最终是奇数个
+			if (cnt[0]+add) >= (add+n+1)/2 && (cnt[2]+add) >= (add+n+1)/2 {
+				return add
+			}
+		} else {
+			// 偶数个
+			if (cnt[0]+add) >= (add+n)/2 && (cnt[2]+add) >= (add+n)/2+1 {
+				return add
+			}
+		}
+	}
+
+	return n + 1
+}

src/codeforces/set0/set1/set100/set160/set166/c/solution_test.go

@@ -0,0 +1,31 @@
+package main
+
+import "testing"
+
+func runSample(t *testing.T, a []int, x int, expect int) {
+	res := solve(a, x)
+	if res != expect {
+		t.Fatalf("Sample expect %d, but got %d", expect, res)
+	}
+}
+
+func TestSample1(t *testing.T) {
+	a := []int{10, 20, 30}
+	x := 10
+	expect := 1
+	runSample(t, a, x, expect)
+}
+
+func TestSample2(t *testing.T) {
+	a := []int{1, 2, 3}
+	x := 4
+	expect := 4
+	runSample(t, a, x, expect)
+}
+
+func TestSample3(t *testing.T) {
+	a := []int{1, 2, 3}
+	x := 3
+	expect := 2
+	runSample(t, a, x, expect)
+}

src/codeforces/set0/set4/set450/set451/c/problem.md

@@ -0,0 +1,27 @@
+There are n games in a football tournament. Three teams are participating in it. Currently k games had already been
+played.
+
+You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your
+friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute
+difference between number of wins of first and second team will be d1 and that of between second and third team will be
+d2.
+
+You don't want any of team win the tournament, that is each team should have the same number of wins after n games.
+That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will
+win this tournament?
+
+Note that outcome of a match can not be a draw, it has to be either win or loss.
+
+### ideas
+
+1. n % 3 = 0 否则不能满足它们有相同的获胜场次
+2. 假设到目前为止，分别获胜 x, y, z 那么有 abs(x - y) = d1, abs(y - z) = d2
+3. x + y + z = k
+4. 假设 y > z
+5. 如果 y > x => y = x + d1 = z + d2
+6. y = k - (x + z)
+7. 2 * y = x + z + d1 + d2
+8. 3 * y = k + d1 + d2 => y
+9. 如果 y < x => y = x - d1 = z + d2 => 2 * y = x + z - d1 + d2 => 3 * y = k - d1 + d2
+10. 如果 y < z
+11. 如果y > x =>

src/codeforces/set0/set4/set450/set451/c/solution.go

@@ -0,0 +1,144 @@
+package main
+
+import (
+	"bufio"
+	"bytes"
+	"fmt"
+	"os"
+)
+
+func readInt(bytes []byte, from int, val *int) int {
+	i := from
+	sign := 1
+	if bytes[i] == '-' {
+		sign = -1
+		i++
+	}
+	tmp := 0
+	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
+		tmp = tmp*10 + int(bytes[i]-'0')
+		i++
+	}
+	*val = tmp * sign
+	return i
+}
+
+func readNum(reader *bufio.Reader) (a int) {
+	bs, _ := reader.ReadBytes('\n')
+	readInt(bs, 0, &a)
+	return
+}
+
+func readTwoNums(reader *bufio.Reader) (a int, b int) {
+	res := readNNums(reader, 2)
+	a, b = res[0], res[1]
+	return
+}
+
+func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
+	res := readNNums(reader, 3)
+	a, b, c = res[0], res[1], res[2]
+	return
+}
+
+func readNNums(reader *bufio.Reader, n int) []int {
+	res := make([]int, n)
+	x := 0
+	bs, _ := reader.ReadBytes('\n')
+	for i := 0; i < n; i++ {
+		for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
+			x++
+		}
+		x = readInt(bs, x, &res[i])
+	}
+	return res
+}
+
+func readUint64(bytes []byte, from int, val *uint64) int {
+	i := from
+
+	var tmp uint64
+	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
+		tmp = tmp*10 + uint64(bytes[i]-'0')
+		i++
+	}
+	*val = tmp
+
+	return i
+}
+
+func main() {
+	reader := bufio.NewReader(os.Stdin)
+	tc := readNum(reader)
+
+	var buf bytes.Buffer
+
+	for tc > 0 {
+		tc--
+		nums := readNNums(reader, 4)
+		res := solve(nums[0], nums[1], nums[2], nums[3])
+		if res {
+			buf.WriteString("yes\n")
+		} else {
+			buf.WriteString("no\n")
+		}
+	}
+
+	fmt.Print(buf.String())
+}
+
+func solve(n int, k int, d1, d2 int) bool {
+	if n%3 != 0 {
+		return false
+	}
+
+	// x <= y  (2种可能性)
+	// z <= y (2种可能性)
+	return solve1(n, k, d1, d2) || solve2(n, k, d1, d2) || solve3(n, k, d1, d2) || solve4(n, k, d1, d2)
+}
+
+func solve1(n int, k int, d1 int, d2 int) bool {
+	// y - x = d1 => y = d1 + x
+	// y - z = d2 => y = d2 + z
+	// y + x + z = k => y = k - (x + z)
+	// 3 * y = k + d1 + d2
+	y := k + d1 + d2
+	if y%3 != 0 {
+		return false
+	}
+	y /= 3
+	return y <= n/3 && y-d1 >= 0 && y-d2 >= 0
+}
+
+func solve2(n int, k int, d1 int, d2 int) bool {
+	// y - x = d1 => y = d1 + x
+	// z - y = d2 => y = z - d2
+	y := k - d2 + d1
+	if y < 0 || y%3 != 0 {
+		return false
+	}
+	y /= 3
+	return y <= n/3 && y >= d1 && y+d2 <= n/3
+}
+
+func solve3(n int, k int, d1 int, d2 int) bool {
+	// x - y = d1 => y = x - d1
+	// y - z = d2 => y = z + d2
+	y := k - d1 + d2
+	if y < 0 || y%3 != 0 {
+		return false
+	}
+	y /= 3
+	return y <= n/3 && y >= d2 && y+d1 <= n/3
+}
+
+func solve4(n int, k int, d1 int, d2 int) bool {
+	// x - y = d1 => y = x - d1
+	// z - y = d2 => y = z - d2
+	y := k - d1 - d2
+	if y < 0 || y%3 != 0 {
+		return false
+	}
+	y /= 3
+	return y <= n/3 && y+d1 <= n/3 && y+d2 <= n/3
+}

src/codeforces/set0/set4/set450/set451/d/problem.md

@@ -0,0 +1,24 @@
+We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. For
+example, "aabba" is good, because after the merging step it will become "aba".
+
+Given a string, you have to find two values:
+
+the number of good substrings of even length;
+the number of good substrings of odd length.
+
+### ideas
+
+1. A substring s[l, r] (1 ≤ l ≤ r ≤ n) of string s = s1s2... sn is string slsl + 1... sr.
+2. 字符串有a/b组成 （0/1)
+3. 如果s[i] = s[i-1]， 那么到i-1的结果，可以直接用在i上，i会和i-1进行合并 （但是长度奇偶性要变化)
+4. 如果s[i] != s[i-1] 那么就需要知道到i-1为止，(假设s[i-1] = 0) 前面是1，且以0开始的（1没有被匹配的）的回文字串的结果
+5. 肯定能确定的一个点是，到目前为止，字符串的pattern必然是 0101 或者1010
+6. 所以dp[0]表示 01, dp[1]表示10, dp[2] 表示101, dp[3]表示010
+7. 然后遇到一个新的0 01 =》 010， 10 =》 10 （但是奇偶性要变）
+8. 101 =》 010 （奇偶性有点麻烦了，如果是偶数个1，和奇数个1）
+9. 010 =》 010 （奇偶性要变)
+9. 不过至少进了一步
+10. 01111 1000111
+11. 如果不考虑merge的情况， 1010101010，其中任意两个0/1之间都是满足回文的条件的
+12. 所以，对于当前0，只需要知道前面和自己相同parity位的0有多少个，加上去，就是奇数长度的计数
+13. 100001