aaa

src/codeforces/set1/set10/set107/set1077/c/problem.md

@@ -0,0 +1,24 @@
+Let's call an array good if there is an element in the array that equals to the sum of all other elements. For example, the array 𝑎=[1,3,3,7]
+ is good because there is the element 𝑎4=7
+ which equals to the sum 1+3+3
+.
+
+You are given an array 𝑎
+ consisting of 𝑛
+ integers. Your task is to print all indices 𝑗
+ of this array such that after removing the 𝑗
+-th element from the array it will be good (let's call such indices nice).
+
+For example, if 𝑎=[8,3,5,2]
+, the nice indices are 1
+ and 4
+:
+
+if you remove 𝑎1
+, the array will look like [3,5,2]
+ and it is good;
+if you remove 𝑎4
+, the array will look like [8,3,5]
+ and it is good.
+You have to consider all removals independently, i. e. remove the element, check if the resulting array is good, and return the element into the array.
+

src/codeforces/set1/set10/set107/set1077/c/solution.go

@@ -0,0 +1,92 @@
+package main
+
+import (
+	"bufio"
+	"fmt"
+	"os"
+)
+
+func main() {
+	reader := bufio.NewReader(os.Stdin)
+
+	n := readNum(reader)
+	a := readNNums(reader, n)
+	res := solve(a)
+
+	fmt.Println(len(res))
+
+	s := fmt.Sprintf("%v", res)
+
+	fmt.Println(s[1 : len(s)-1])
+}
+
+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 solve(a []int) []int {
+	freq := make(map[int]int)
+
+	var sum int
+	for _, num := range a {
+		sum += num
+		freq[num]++
+	}
+
+	var res []int
+	for i, num := range a {
+		freq[num]--
+		tmp := sum - num
+		if tmp%2 == 0 {
+			tmp /= 2
+			if freq[tmp] > 0 {
+				res = append(res, i+1)
+			}
+		}
+		freq[num]++
+	}
+	return res
+}

src/codeforces/set1/set10/set107/set1077/c/solution_test.go

@@ -0,0 +1,28 @@
+package main
+
+import (
+	"reflect"
+	"sort"
+	"testing"
+)
+
+func runSample(t *testing.T, a []int, expect []int) {
+	res := solve(a)
+
+	sort.Ints(expect)
+
+	if !reflect.DeepEqual(res, expect) {
+		t.Fatalf("Sample expect %v, but got %v", expect, res)
+	}
+}
+
+func TestSample1(t *testing.T) {
+	a := []int{2, 5, 1, 2, 2}
+	expect := []int{4, 1, 5}
+	runSample(t, a, expect)
+}
+
+func TestSample2(t *testing.T) {
+	a := []int{2, 1, 2, 4, 3}
+	runSample(t, a, nil)
+}

src/codeforces/set1/set19/set195/set1950/e/problem.md

@@ -0,0 +1,24 @@
+You are given a string 𝑠
+ of length 𝑛
+ consisting of lowercase Latin characters. Find the length of the shortest string 𝑘
+ such that several (possibly one) copies of 𝑘
+ can be concatenated together to form a string with the same length as 𝑠
+ and, at most, one different character.
+
+More formally, find the length of the shortest string 𝑘
+ such that 𝑐=𝑘+⋯+𝑘𝑥 times
+ for some positive integer 𝑥
+, strings 𝑠
+ and 𝑐
+ has the same length and 𝑐𝑖≠𝑠𝑖
+ for at most one 𝑖
+ (i.e. there exist 0
+ or 1
+ such positions).
+
+### ideas
+1. k = n时，是成立的
+2. k肯定是n个一个因子
+3. 然后在给定k的情况下，可以在O(n)的时间内检查？
+4. 是可以的
+5. 

src/codeforces/set1/set19/set195/set1950/e/solution.go

@@ -0,0 +1,141 @@
+package main
+
+import (
+	"bufio"
+	"bytes"
+	"fmt"
+	"os"
+)
+
+func main() {
+	reader := bufio.NewReader(os.Stdin)
+
+	tc := readNum(reader)
+
+	var buf bytes.Buffer
+
+	for tc > 0 {
+		tc--
+		readNum(reader)
+		s := readString(reader)
+		res := solve(s)
+		buf.WriteString(fmt.Sprintf("%d\n", res))
+	}
+
+	fmt.Print(buf.String())
+}
+
+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 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 solve(s string) int {
+	n := len(s)
+	flag := make([]bool, n)
+	check := func(k int) bool {
+		for i := 0; i < k; i++ {
+			flag[i] = false
+		}
+		for i := 0; i < n; i++ {
+			if s[i] != s[i%k] {
+				flag[i%k] = true
+			}
+		}
+		diff := -1
+		for i := 0; i < k; i++ {
+			if flag[i] {
+				if diff >= 0 {
+					return false
+				}
+				diff = i
+			}
+		}
+		if diff < 0 {
+			return true
+		}
+		if n/k == 2 {
+			return true
+		}
+
+		var expect byte
+
+		if s[diff] == s[diff+k] || s[diff] == s[diff+2*k] {
+			// s[diff]是多的那方
+			expect = s[diff]
+		} else {
+			expect = s[diff+k]
+		}
+		var cnt int
+		for i := diff; i < n; i += k {
+			if s[i] != expect {
+				cnt++
+			}
+		}
+		return cnt == 1
+	}
+	res := n
+	for k := 1; k <= n/k; k++ {
+		if n%k == 0 {
+			if check(k) {
+				res = min(res, k)
+			}
+			if k*k != n && check(n/k) {
+				res = min(res, n/k)
+			}
+		}
+	}
+	return res
+}

src/codeforces/set1/set19/set195/set1950/e/solution_test.go

@@ -0,0 +1,29 @@
+package main
+
+import "testing"
+
+func runSample(t *testing.T, s string, expect int) {
+	res := solve(s)
+
+	if res != expect {
+		t.Fatalf("Sample expect %d, but got %d", expect, res)
+	}
+}
+
+func TestSample1(t *testing.T) {
+	s := "abaa"
+	expect := 1
+	runSample(t, s, expect)
+}
+
+func TestSample2(t *testing.T) {
+	s := "slavicgslavic"
+	expect := 13
+	runSample(t, s, expect)
+}
+
+func TestSample3(t *testing.T) {
+	s := "hshahaha"
+	expect := 2
+	runSample(t, s, expect)
+}

src/codeforces/set1/set19/set195/set1950/f/problem.md

@@ -0,0 +1,24 @@
+Find the minimum height of a rooted tree†
+ with 𝑎+𝑏+𝑐
+ vertices that satisfies the following conditions:
+
+𝑎
+ vertices have exactly 2
+ children,
+𝑏
+ vertices have exactly 1
+ child, and
+𝑐
+ vertices have exactly 0
+ children.
+If no such tree exists, you should report it.
+
+
+### ideas
+1. 显然c是叶子节点的个数
+2. b是只有一个子child的内部节点
+3. 它们只能排成一条线
+4. (b + c - 1) / c + 1 是它们的最低高度
+5. 先假设能整除，此时可以认为仍然有c的叶子节点
+6. 然后就是a个有两个节点的内部节点。这样的节点可以组成一个完全二叉树, 
+7. a <= c,