42-2017蓝桥杯b java

#include <iostream>
#include <algorithm>
using namespace std;int main(){
int n, m;
int a[15];
for(int i = 0; i < 9; i++){
a[i] = i + 1;
}
int cnt = 0; do{
int c1 = a[0] + a[1] + a[2] + a[3];
int c2 = a[3] + a[4] + a[5] + a[6];
int c3 = a[6] + a[7] + a[8] + a[0];
if(c1 == c2 && c2 == c3)
cnt++;
// cout << a[0] << "," << a[1] << "," << a[2] << endl;
}while(next_permutation(a, a + 9));
cout << "cnt: " << cnt << endl;/*
全排列库函数：
next_permutation:
1.必须是有序的从小大排好序的才行；
2.参数：数组名，数组名+排序的长度 */ return 0;
}

import java.util.Scanner;public class Main {
public static int[] visit = new int[10];
public static int[] a = new int[10];
public static int n;
public static int all = 0;
public static int ys = 0;public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
n = cin.nextInt();
// long ans = 1;
// for(int i = 1; i <= 9; i++) { //九的全排列：362880
// ans *= i;
// }
// System.out.println(ans);
dfs(0);
System.out.println(all); //362880
System.out.println(ys / 6);}
public static void dfs(int cnt) {
if(cnt == n) {
all++;
//打印所有全排列
//for(int i = 0; i < n; i++) {
////System.out.print(box[i]);
//System.out.printf("%5d", box[i]);
//}
//System.out.println();
//判断是否和要求
int x = a[0] + a[1] + a[2] + a[3];
int x1 = a[3] + a[4] + a[5] + a[6];
int x2 = a[6] + a[7] + a[8] + a[0];
if(x == x1 && x == x2) {
//System.out.println();
ys++;
}
return ;
}
for(int i = 1; i <= n; i++) {
if(visit[i] == 0) {
a[cnt] = i;
visit[i] = 1;
dfs(cnt + 1);
visit[i] = 0;
}
}
}
}

import java.util.Arrays;
import java.util.Scanner;public class Main {
public static int[] visit = new int[10];
public static int n;
public static double[][] a = new double[40][40];public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
// n = cin.nextInt();
n = 29;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= i; j++) {
long x = cin.nextLong();
a[i][j] += x;
a[i+1][j] += a[i][j] / 2.0;
a[i+1][j+1] += a[i][j] / 2.0;
}
}
//打印处理后的数据，验证前几个看是否符合预期
for(int i = 1; i <= n + 1; i++) {
for(int j = 1; j <= i; j++) {
System.out.print(a[i][j] + " ");
}
System.out.println();
}
System.out.println("============"); //自己寻找最大最小值：
double min = 0x3f3f3f * 1.0;
double max = -1;
for(int i = 1; i <= 30; i++) {
if(a[30][i] > max) max = a[30][i];
if(a[30][i] < min) min = a[30][i];
}
System.out.println("min" + min + "; man " + max); //用函数排序找最大最小值：
Arrays.sort(a[30]);
for(int i = 0; i < 40; i++) {
System.out.print(a[30][i] + " ");
}
System.out.println(); System.out.println("min" + a[30][10] + "; man " + a[30][39]);
double ans = (a[30][39] / a[30][10]) * 2086458231;
System.out.println(ans);
System.out.println((long)ans);// 7.2665192664E10
// 72665192664
}
}

public class Main
{
static int len(int x){
if(x<10) return 1;
return len(x/10)+1;
} // 取x的第k位数字
static int f(int x, int k){
if(len(x)-k==0)
return x%10;
return __________; //填空
} public static void main(String[] args)
{
int x = 23513;
//System.out.println(len(x));
System.out.println(f(x,3));
}
}

/*
6.最大公共子串最大公共子串长度问题就是：
求两个串的所有子串中能够匹配上的最大长度是多少。比如："abcdkkk" 和 "baabcdadabc"，
可以找到的最长的公共子串是"abcd",所以最大公共子串长度为4。下面的程序是采用矩阵法进行求解的，这对串的规模不大的情况还是比较有效的解法。
*/
public class Main
{
static int f(String s1, String s2)
{
char[] c1 = s1.toCharArray();
char[] c2 = s2.toCharArray();int[][] a = new int[c1.length+1][c2.length+1];int max = 0;
for(int i=1; i<a.length; i++){
for(int j=1; j<a[i].length; j++){
if(c1[i-1]==c2[j-1]) {
; //填空
if(a[i][j] > max) max = a[i][j];
}
}
}
return max;
}public static void main(String[] args){
int n = f("abcdkkk", "baabcdadabc");
System.out.println(n);
}
}

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;public class Main{
//public static int [] a = new int[20];public static void main(String[] args){Scanner cin = new Scanner(System.in);
List<Data> list = new ArrayList<Data>();
int x, y, z; //AA/BB/CC
String str = cin.next();
x = (int)(str.charAt(0)-'0') * 10 + (int)(str.charAt(1)-'0');
y = (int)(str.charAt(3)-'0') * 10 + (int)(str.charAt(4)-'0');
z = (int)(str.charAt(6)-'0') * 10 + (int)(str.charAt(7)-'0');
int xx = x, yy = y, zz = z;//System.out.println(x + "; " + y + "; " + z);
if((x >= 0 && x <= 59) || (x >= 60 && x <= 99)) {
int year = 0;
if(x >= 0 && x <= 59 ) {
year = x + 2000;
}
else {
year = x + 1900;
}
int run = 0;
if(year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))
run = 1;
//System.out.println( year + " test+ " + run);
if(y >= 1 && y <= 12) {
if(y == 1 || y == 3 || y == 5 || y == 7 || y == 8 || y == 10 || y == 12) {
if(z >= 1 && z <= 31) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 0) {
if(z >= 1 && z <= 28) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 1) {
if(z >= 1 && z <= 29) {
list.add(new Data(x,y,z));
}
}
else {
if(z >= 1 && z <= 30) {
list.add(new Data(x,y,z));
}
}
}
}
x = zz; y = yy; z = xx;
if((x >= 0 && x <= 59) || (x >= 60 && x <= 99)) {
int year = 0;
if(x >= 0 && x <= 59 ) {
year = x + 2000;
}
else {
year = x + 1900;
}
int run = 0;
if(year % 400 == 0 || (year % 100 == 1 && year % 4 == 0))
run = 1;
if(y >= 1 && y <= 12) {
if(y == 1 || y == 3 || y == 5 || y == 7 || y == 8 || y == 10 || y == 12) {
if(z >= 1 && z <= 31) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 0) {
if(z >= 1 && z <= 28) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 1) {
if(z >= 1 && z <= 29) {
list.add(new Data(x,y,z));
}
}
else {
if(z >= 1 && z <= 30) {
list.add(new Data(x,y,z));
}
}
}
}
x = zz; y = xx; z = yy;
if((x >= 0 && x <= 59) || (x >= 60 && x <= 99)) {
int year = 0;
if(x >= 0 && x <= 59 ) {
year = x + 2000;
}
else {
year = x + 1900;
}
int run = 0;
if(year % 400 == 0 || (year % 100 == 1 && year % 4 == 0))
run = 1;
if(y >= 1 && y <= 12) {
if(y == 1 || y == 3 || y == 5 || y == 7 || y == 8 || y == 10 || y == 12) {
if(z >= 1 && z <= 31) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 0) {
if(z >= 1 && z <= 28) {
list.add(new Data(x,y,z));
}
}
else if (y == 2 && run == 1) {
if(z >= 1 && z <= 29) {
list.add(new Data(x,y,z));
}
}
else {
if(z >= 1 && z <= 30) {
list.add(new Data(x,y,z));
}
}
}
}
System.out.println(list.toString());
Collections.sort(list, new Comparator<Data>() {
public int compare(Data y, Data x) {
if(y.year < x.year) {
return -1;
}
else if(y.year > x.year){
return 1;
}
else {
if(y.math < x.math) {
return -1;
}
else if(y.math > x.math){
return 1;
}
else {
if(y.day < x.day) {
return -1;
}
else if(y.day > x.day){
return 1;
}
else {
return 0;
}
}
}
}
});for(Data d : list) {
if(d.year >= 60) {
System.out.print("19" + d.year);
}
else {
System.out.print("20");
if(d.year >= 10) {
System.out.print(d.year);
}
else {
System.out.print("0" + d.year);
}
if(d.math >= 10) {
System.out.print("/"+ d.math);
}
else {
System.out.print("/0"+ d.math);
}
if(d.day >= 10) {
System.out.print("/"+ d.day);
}
else {
System.out.print("/0"+ d.day);
}
}
System.out.println();
}}}class Data{
int year;
int math;
int day;
Data(int y, int m, int d){
this.year = y;
this.math = m;
this.day = d;
}
@Override
public String toString() {
return "Data [year=" + year + ", math=" + math + ", day=" + day + "]";
}
}
/*
02/03/04
06/05/04
04/02/29
05/02/29
*/

import java.util.Scanner;public class Main{
public static int a[] = new int[104];
public static int dp[] = new int[10008];public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n;
n = cin.nextInt();
int flag = 1;
for(int i = 0; i < n; i++) {
a[i] = cin.nextInt();
}
flag = a[0];
for(int i = 1; i < n; i++) {
flag = gcd(flag, a[i]);
if(flag == 1) { // 如果有互质的则一定是有限个了
break;
}
}
if(flag != 1) {
System.err.println("INF");
return ;
}dp[0] = 1; // 标记为可以凑出
for(int i = 0; i < n; i++) {
for(int j = a[i]; j <= 10004; j++) {
if(dp[j - a[i]] == 1) { //如果可以凑出来，进行标记
dp[j] = 1;//能组合的全是2的倍数了，非2的倍数就取不到了
}
}
}
int ans = 0;
for(int i = 0; i < 10004; i++) {
if(dp[i] == 0) {
ans++;
}
}
System.out.println(ans);
}public static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
} }

import java.util.Scanner;public class Main{
public static int w[] = new int[100005];
public static int h[] = new int[100005];
public static int n;
public static int k;public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
n = cin.nextInt();
k = cin.nextInt();int max = -1;
for(int i = 0; i < n; i++) {
w[i] = cin.nextInt();
h[i] = cin.nextInt();
max = Math.max(max, w[i]);
max = Math.max(max, h[i]);
}int left = 1, right = 100000;
// 或者用最大值：max
right = max;
int mid;
// 二分查找可能的值：搜索空间：1 ：100000
while(left < right) {
mid = left + (right - left) / 2;
if(slove(mid) == 1) {
left = mid + 1;
}
else {
right = mid - 1;
}
//System.out.println("--" + mid);
}
System.out.println(left - 1);
}public static int slove(int s) {
int cnt = 0;
for(int i = 0; i < n; i++) {
cnt += w[i] / s * h[i] / s;
}
if(cnt >= k) {
return 1;
}
return 0;
}
}

import java.util.Scanner;//0<N,V≤1000 表示物品数量和背包容积。
//0<vi,wi≤1000 体积和 价值
public class Main{
public static int a[] = new int[100005];
public static int cnt[] = new int[100005];public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n, k;
n = cin.nextInt();
k = cin.nextInt();int ans = 0;
for(int i = 1; i <= n; i++) {
a[i] = cin.nextInt();
a[i] = (a[i - 1] + a[i]) % k;
ans += cnt[a[i]];
cnt[a[i]]++;
}//for(int i = 0 ; i < k; i++) {
//System.out.println(cnt[i]);
//}System.out.println(ans + cnt[0]);}}