Determine the number of dreams
Title Description
Dream defined number, the number of each of the squares and has add operation occurs when the number of cycles in a dream was
enter a positive integer, it is determined whether the number is a dream
Example: 19
1 ^ 9 ^ 2 + 2 = 82
. 8 ^ 2 + 2 ^ 2 = 68
^ 6 2 + 2 = 100 8 ^
1 ^ 0 ^ 2 + 2 + 2 ^ 0 = 1
i.e. the number 19 is a dream
coding
// TODORye escape
Title Description
Master lived in a village, an avalanche day Master running speed of 13m / s, using magic 1s move 50m, but will spend 10 Magic
Master initial mana M, village distance safely is S, the time avalanche arrival T, the Master recovery magic rate of 4:00 / s, but only standing still can recover
Entry
Input line, comprising three non-negative integers M, S, T
Export
Two output lines
of the first line of output "Yes", "No", indicates whether the safe arrival
or non-arrival of the second row reaches the output time of arrival of the maximum distance
coding
// 代码为自行编写，可能存在 Bug
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int m = scn.nextInt(), s = scn.nextInt(), t = scn.nextInt();
// count：初始魔法值能用几次 tAns：记录使用的时间 sAns：记录走过的路程 flag：此处标记剩余魔法值是否为 4 的倍数
int count = m/10, sAns = 0, tAns = 0, tt = 0, flag = (10-m%10)%10%4 == 0 ? 0 : 1;
if(m%10 > 1 && m%10/4 + flag <= t) { // tt记录下恢复到至少 10 点魔法值所需时间
tt = m%10 / 4 + flag;
}
flag = s%50 == 0 ? 0 : 1; // 此处标记路程是否为 50 的倍数
count = Math.min(count, Math.min(s/50+flag,t));
sAns += count * 50;
tAns += count;
if(tt != 0 && tt+1 <= t-tAns && tt*13 + sAns < s) { // 存在多余魔法值 未被淹没 使用等待的时间直接走不会到达目的地
sAns += 50;
tAns += tt + 1;
}
flag = (s-sAns)%13 == 0 ? 0 : 1; // 此处标记剩余路程为 13 的倍数
count = Math.min((s-sAns)/13+flag, (t-tAns));
sAns += count * 13;
tAns += count;
if(sAns < s) {
System.out.println("No");
System.out.println(sAns);
} else {
System.out.println("Yes");
System.out.println(tAns);
}
}
}Travel Route
Title Description
There are cities, M N Block route, which now want to go to Marco Polo R cities, please help him develop a shortest route.
Course for undirected edges, to ensure that no heavy side
Entry
The first input line N, M, R (2 < = N <= 200,1 <= M <= 500,2 <= R <= min (22, N)
a second input line number R cities
M rows after per line x, y, z represents the distance between the cities and urban y z x m (z <= 10000)
Export
The shortest distance route
coding
// TODOTo Be Continue...