/*因为 1<<32 不能用int来接收，不然这个题应该是考察溢出的问题，用long来做完全没挑战性*/
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int group = sc.nextInt();
		for(int i = 1; i <= group; i++){
			long numA = sc.nextLong();
			long numB = sc.nextLong();
			long numC = sc.nextLong();
			String res="Case #"+i+":"+" ";
			res += numA + numB > numC?"true":"false";
			System.out.println(res);
		}
	}
}

python 4行解法：

a=int(input())
for i in range(a):
    b,c,d=map(int,input().split())
    print("Case #"+str(i+1)+": true" if b+c>d else "Case #"+str(i+1)+": false")

/**
 * A+B和C (15)
 * 时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)
 * 题目描述
 * 给定区间[-2的31次方, 2的31次方]内的3个整数A、B和C，请判断A+B是否大于C。
 * 输入描述:
 * 输入第1行给出正整数T(<=10)，是测试用例的个数。随后给出T组测试用例，每组占一行，顺序给出A、B和C。整数间以空格分隔。
 * 输出描述:
 * 对每组测试用例，在一行中输出“Case #X: true”如果A+B>C，否则输出“Case #X: false”，其中X是测试用例的编号（从1开始）。
 * 输入例子:
 * 4
 * 1 2 3
 * 2 3 4
 * 2147483647 0 2147483646
 * 0 -2147483648 -2147483647
 * 输出例子:
 * Case #1: false
 * Case #2: true
 * Case #3: true
 * Case #4: false
 *
 * @author shijiacheng
 */
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            BigInteger A = sc.nextBigInteger();
            BigInteger B = sc.nextBigInteger();
            BigInteger C = sc.nextBigInteger();
            BigInteger sum = A.add(B);
            if (sum.compareTo(C) > 0) {
                System.out.println("Case #" + (i + 1) + ": true");
            } else {
                System.out.println("Case #" + (i + 1) + ": false");
            }
        }
    }
}

 就跟我说
1
2147483648 2147483648 2147483648
过不去 然而我这能过去啊... 什么鬼

import java.util.Scanner;
public class Main{	
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		boolean[] result = new boolean[n];
		for(int i = 0; i < n; i++)
		{
			long a = sc.nextLong();
			long b = sc.nextLong();
			long c = sc.nextLong();
			if((c-a)<b) result[i] = true;
			else result[i] = false;
		}
		for(int i = 0; i < n; i++)
			System.out.println("Case #" + (i+1) + ": "+result[i]);
	}
}

import java.util.ArrayList;
import java.util.Scanner;

public class Test1 {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		ArrayList<String> list = new ArrayList<String>();
		long a,b,c;
		int i=1;
		int count = scan.nextInt();
		while(count > 0){
			a = scan.nextLong();
			b = scan.nextLong();
			c = scan.nextLong();
			if(a+b>c)list.add("true");
			else list.add("false");
			count --;
		}
		/*
		 * case #1: false
		 */
		for(String str:list){
			System.out.println("Case #"+i+": "+str);
			i++;
		}
		
			
	}

}

#include<iostream>
using namespace std;
int main(void)
{
    int T;
    cin >> T;
    long int **pt = new long int*[T];
    for (int i = 0; i<T; i++)
    {
        pt[i] = new long int[3];
        cin >> pt[i][0] >> pt[i][1] >> pt[i][2];
    }
    for (int i = 0; i < T; i++)
        if (pt[i][0] + pt[i][1] > pt[i][2])
            cout << "Case #" << i+1 << ": true" << endl;
        else
            cout << "Case #" << i+1 << ": false" << endl;
    return 0;
}

#include <stdio.h>
#include <stdlib.h>

int comp(long int a,long int b,long int c)
{
	int r =(a+b)>c?(1):(0);
	return r;
}

int main()
{
	int num = 0,i=0;
	long int a,b,c;
	scanf("%d",&num);
	int result[num];
	for(i=0;i<num;i++){
		scanf("%ld%ld%ld",&a,&b,&c);
		result[i] = comp(a,b,c);
	}
	for(i=0;i<num;i++){
		if(result[i]==1)
			printf("Case #%d: true\n",i+1);
		else
			printf("Case #%d: false\n",i+1);
	}
	return 0;
}
	

• 按照题目要求，先接收输入的测试组数，再输入每一组数据时，进行调用比较函数判断并将结果存于一个暂时的数组中，做好对应关系。等全部输入完后，再将原先存好的数组数据判断，输出结果，如此，并可实现题目的一串输入，一串结果输出。

import java.util.*;
//简单题
public class Main {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int i =0;
        while(t-->0){
            i++;
            long a = in.nextLong();
            long b = in.nextLong();
            long c = in.nextLong();
            System.out.println("Case #"+i+": "+(a+b>c));
        }
    }
} 

import java.util.Scanner;

public class Main {

public static void main(String[] args) {
	    int p = 0;
		//System.out.println("Type T that 0 < T <= 10 :");
		Scanner _scanner = new Scanner(System.in) ;
		int T = _scanner.nextInt() ;
		
		while (T <= 0 || T > 10) {
			System.out.println("retype T that 0 < T <= 10 :");
			Scanner _scanner1 = new Scanner(System.in) ;
			T = _scanner1.nextInt() ;
			}
		
		//System.out.println("输入" + T + "组测试用例，每组A B C三个数，并用空格隔开：");
		
		long[][] _2darray = new long[T][3] ;
		
		for (int i = 0; i < T; i++) {
			for (int j = 0; j < 3; j++) {
				_2darray[i][j] = _scanner.nextLong() ;
			}
		}
		
		  for (int i = 0; i < T; i++) {
			  
			if (_2darray[i][0] + _2darray[i][1] > _2darray[i][2]) {
				System.out.println("Case #" + ++p + ": " + "true");
			} else {
				System.out.println("Case #" + ++p + ": " + "false");
			}
		}
		
	}

}

s = input()
for i in range(1,s+1):
    num = raw_input().split(' ')
    if int(num[0])+int(num[1])>int(num[2]):
        print 'Case #'+str(i)+': true'
    else:
        print 'Case #'+str(i)+': false'

分析

1. 因为在给定区间 判断，该区间右端点在int范围已经溢出，且也在int范围是溢出的，说明加数，和均会溢出。所以不能使用int类型存储输入数据，需要用long long类型来保证计算不会溢出。

/*
 * app=PAT-Basic lang=c++
 * https://pintia.cn/problem-sets/994805260223102976/problems/994805312417021952
 */
#include <cstdio>
int main()
{
    int T;
    long long A, B, C,sum;
    scanf("%d",&T);
    for (int i = 0; i < T;i++){
        scanf("%lld%lld%lld",&A,&B,&C);
        sum = A + B;
        if (sum > C){
            printf("Case #%d: true\n",i + 1);
        }
        else{
            printf("Case #%d: false\n", i + 1);
        }
    }
    return 0;
}

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        byte T = in.nextByte();
        for (int i = 1; i <= T; i++)
            System.out.println("Case #" + i + ": " + (in.nextLong() + in.nextLong() > in.nextLong()));
    }
}

import   java.util.Scanner;