【ACM】P1000、P10001、P1002、P1003代码演示

（p1000） 问题描述（A+B Problem）：

/***
* A + B Problem（解决A+B简单问题）
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 658688    Accepted Submission(s): 205290

Problem Description

Calculate A + B.//计算A+B

Input  //输入
Each line will contain two integers A and B. Process to end of file.

Output
For each case, output A + B in one line.

Sample Input

1 1

Sample Output

2

Author
HDOJ
*@author <a"283505495@qq.com">lxd</a>
*@version 1.0 2017-4-26 下午1:53:07
*@fileName P1000.java
*/

代码演示：

package ac;
import java.util.Scanner;
public class P1000 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int a=sc.nextInt();
int b=sc.nextInt();
int sum=a+b;
System.out.println(sum);
}
}
}

运行结果：

p1001问题描述(Sum Problem)：

/**
* Sum Problem（求和问题）
Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 470318    Accepted Submission(s): 118540

Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.（求解1+....+n的和）

Input
The input will consist of a series of integers n, one integer per line.

Output
For each case, output SUM(n) in one line, followed by a blank line.
32位的整数 也许n的值特别大
You may assume the result will be in the range of 32-bit signed integer.

Sample Input

1
100

Sample Output

1

5050
*@author <a"283505495@qq.com">lxd</a>
*@version 1.0 2017-4-26 下午2:09:34
*@fileName p1001.java
*/

代码演示：

package ac;
import java.util.Scanner;
public class p1001 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
int sum=0;
for(int i=1;i<=n;i++){
sum+=i;
}
System.out.println(sum);
System.out.println();
}
}
}

运行结果：

p1002问题描述(A + B Problem II（A+B问题2 —–利用大数进行解决）)：

/**
* A + B Problem II（A+B问题2 -----利用大数进行解决）
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 356144    Accepted Submission(s): 69087

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input

（情况个数T）
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases.
（这两个A、B数可能非常大）
Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large,
（比32位整数还要大）
that means you should not process them by using 32-bit integer.
（整数长度不要超过1000）
You may assume the length of each integer will not exceed 1000.

Output
For each test case, you should output two lines.
The first line is "Case #:", # means the number of the test case.
he second line is the an equation "A + B = Sum", Sum means the result of A + B.
Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
*@author <a"283505495@qq.com">lxd</a>
*@version 1.0 2017-4-26 下午2:16:06
*@fileName P1002.java
*/

代码演示：

package ac;

import java.math.BigInteger;
import java.util.Scanner;

public class P1002 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
/**
* 1、java.util.math中的BigInteger类，用来封装大数，可以对大数进行基本的操作。
*/
int t=sc.nextInt();
for(int i=0;i<t;i++){
BigInteger a=new BigInteger(sc.next());
BigInteger b=new BigInteger(sc.next());
BigInteger sum=a.add(b);
System.out.println("Case "+(i+1)+":");
/**
* a、b、sum必须转换为toString
*/
System.out.println(a.toString()+" + "+b.toString()+" = "+sum.toString());
if(i!=t-1){
System.out.println();
}
}
}
}

运行结果：

p1003问题描述(Max Sum（求一个无序序列的最大和）)：

/**
* Max Sum（求一个无序序列的最大和）
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242414    Accepted Submission(s): 57233

Problem Description
给定一个序列值，计算出此序列的最大和
Given a sequence a[1],a[2],a[3]......a
, your job is to calculate the max sum of a sub-sequence.
For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases.
Then T lines follow, each line starts with a number N(1<=N<=100000),
then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case.
最大和、起始位置、结束位置
The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence,
the end position of the sub-sequence. If there are more than one result, output the first one.
Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

*@author <a"283505495@qq.com">lxd</a>
*@version 1.0 2017-4-26 下午3:02:59
*@fileName P1003.java
*/

代码演示：