【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): 205290Problem DescriptionCalculate A + B.//计算A+BInput  //输入Each line will contain two integers A and B. Process to end of file.OutputFor each case, output A + B in one line.Sample Input1 1Sample Output2AuthorHDOJ *@author <a"[e-mail]">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): 118540Problem DescriptionHey, 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）InputThe input will consist of a series of integers n, one integer per line.OutputFor 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 Input1100Sample Output15050 *@author <a"[e-mail]">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): 69087Problem DescriptionI 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.OutputFor 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 Input21 2112233445566778899 998877665544332211Sample OutputCase 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110 *@author <a"[e-mail]">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): 57233Problem 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[n], 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.InputThe 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).OutputFor 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 Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5Sample OutputCase 1:14 1 4Case 2:7 1 6 *@author <a"[e-mail]">lxd</a> *@version 1.0 2017-4-26 下午3:02:59 *@fileName P1003.java */

代码演示：