在商城的某个位置有一个商品列表，该列表是由L1、L2两个子列表拼接而成。当用户浏览并翻页时，需要从列表L1、L2中获取商品进行展示。展示规则如下：
1. 用户可以进行多次翻页，用offset表示用户在之前页面已经浏览的商品数量，比如offset为4，表示用户已经看了4个商品
2. n表示当前页面需要展示的商品数量
3. 展示商品时首先使用列表L1，如果列表L1长度不够，再从列表L2中选取商品
4. 从列表L2中补全商品时，也可能存在数量不足的情况
请根据上述规则，计算列表L1和L2中哪些商品在当前页面被展示了
每个测试输入包含1个测试用例，包含四个整数，分别表示偏移量offset、元素数量n，列表L1的长度l1，列表L2的长度l2。
在一行内输出四个整数分别表示L1和L2的区间start1，end1，start2，end2，每个数字之间有一个空格。 注意，区间段使用半开半闭区间表示，即包含起点，不包含终点。如果某个列表的区间为空，使用[0, 0)表示，如果某个列表被跳过，使用[len, len)表示，len表示列表的长度。
2 4 4 4 1 2 4 4 4 1 3 3
2 4 0 2 1 3 0 0 3 3 1 2
#include <algorithm>
#include <cstdio>
using namespace std;
int main() {
int offset, n, l1, l2;
while (~scanf("%d%d%d%d", &offset, &n, &l1, &l2))
printf("%d %d %d %d\n", min(l1, offset), min(l1, offset+n), min(
max(offset-l1, 0), l2), min(max(offset+n-l1, 0), l2));
return 0;
}
认真仔细读题 分析所有可能情况 这题没啥难度 就是要考虑周全 合理分类
#include<iostream>
#include<algorithm>
using namespace std;
int main(){
int off, n, l1, l2;
int start1, end1, start2, end2;
while(scanf("%d%d%d%d", &off, &n, &l1, &l2) != EOF){
if(off >= l1 + l2){//l1和l2都被跳过
start1 = l1, end1 = l1;
start2 = l2, end2 = l2;
}
else if(off >= l1){//l1被跳过
start1 = l1, end1 = l1;
start2 = off - l1, end2 = min(l2, off + n - l1);
}
else if(off + n <= l1){//l2为空
start2 = 0, end2 = 0;
start1 = off, end1 = off + n;
}
else{//两边都有的情况
start1 = off, end1 = l1;
start2 = 0, end2 = min(l2, off + n - l1);
}
printf("%d %d %d %d\n", start1, end1, start2, end2);
}
return 0;
}
/*
2 4 4 4
1 2 4 4
4 1 3 3
*/
#include <iostream>
using namespace std;
int main()
{
int offset, n, l1, l2;
while(cin >> offset >> n >> l1 >> l2) {
if(offset < l1) {
cout << offset << " " << min(offset + n, l1) << " ";
if(offset + n - l1 < 0) {
cout << "0 0" << endl;
} else {
cout << min(offset + n - l1, 0) << " " << min(max(offset + n - l1, 0), l2) << endl;
}
} else if(offset >= l1) {
cout << l1 << " " << l1 << " ";
cout << min(offset - l1, l2) << " " << min(offset + n - l1, l2) << endl;
}
}
return 0;
}
while True:
try:
offset, n, len1, len2 = map(int, input().strip().split())
if len1 - offset > 0:
# 第一个列表没有翻过
start1 = offset # 第一个列表的起始位置
end1 = min(start1 + n, len1) # 第一个列表的结束位置
if end1 - start1 >= n:
# 第一个列表够补的话就不需要第二个列表的商品
start2, end2 = 0, 0
else:
# 不够就需要从前往后取第二个列表的商品
start2 = 0
end2 = min(start2 + n - (end1 - start1), len2)
print(f"{start1} {end1} {start2} {end2}")
elif len1 == offset:
# 第一个列表刚好翻过
start1 = end1 = len1
start2 = 0
end2 = min(start2 + n, len2)
print(f"{start1} {end1} {start2} {end2}")
else:
# 第一个列表翻完了，第二个列表也看过了一些
start1 = end1 = len1
start2 = min(offset - len1, len2)
if start2 == len2:
# 第二个列表也翻完了
end2 = len2
else:
end2 = min(start2 + n, len2)
print(f"{start1} {end1} {start2} {end2}")
except:
break 
#include <bits/stdc++.h>
using namespace std;
int main(){
int offset, n, l1, l2, s1, s2, e1, e2;
while(scanf("%d%d%d%d", &offset, &n, &l1, &l2) != EOF){
int m = offset+n;
if(offset <= l1){
s1 = offset;
if(m <= l1){
e1 = m;
s2 = e2 = 0;
}else{
e1 = l1;
s2 = 0;
e2 = min(m-l1, l2);
}
}else if(offset <= l1+l2){
s1 = e1 = l1;
s2 = offset-l1;
e2 = min(m-l1, l2);
}else{
s1 = e1 = l1;
s2 = e2 = l2;
}
printf("%d %d %d %d\n", s1, e1, s2, e2);
}
return 0;
} 
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int offset=sc.nextInt();
int n=sc.nextInt();
int l1=sc.nextInt();
int l2=sc.nextInt();
showList(offset,n,l1,l2);
}
public static void showList(int offset,int n,int l1,int l2){
/* 分情况讨论：展示的长度始终为n，范围在offset——offset+n之间，
* 对该范围进行分情况讨论。难点在于将题中描述提取出自己需要的信息，
* 并进行转化。
*/
if(offset <= l1 + l2){
// 表明起点在L2表中
if (offset >= l1){
if (offset+n >= l1 + l2){
// 表明总长度大于L1，L2列表长度和
System.out.print(l1 + " " + l1 + " " + (offset-l1) + " " + l2);
} else {
// 表明长度小于L1，L2列表长度和
System.out.print(l1 + " " + l1 + " " + (offset-l1) + " " + (offset+n-l1));
}
} else if(offset+n<=l1){
//表明展示的数据全部来自L1
System.out.print(offset + " " + (offset+n) + " 0 0");
} else if (offset<l1){
// 来自两表
System.out.print(offset + " " + l1 + " 0 " + (offset+n-l1));
}
} else {
// 表示两表跳过
System.out.print(l1+" "+l1+" "+l2+" "+l2);
}
}
} 
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int offset=sc.nextInt();
int n=sc.nextInt();
int len1=sc.nextInt();
int len2=sc.nextInt();
int[] res=getResult(offset,n,len1,len2);
for(int i=0;i<res.length;i++){
if(i==res.length-1){
System.out.println(res[i]);
}else{
System.out.print(res[i]+" ");
}
}
}
}
private static int[] getResult(int offset, int n, int len1, int len2) {
int[] res=new int[4];
//start1
res[0]=len1-offset>0?offset:len1;
//end1
res[1]=len1-offset-n>0?offset+n:len1;
//start2
int k=offset-len1;
res[2]=k>0?len2-k>0?k:len2:0;
//end2
int x=offset+n-len1;
res[3]=x>0?len2-x>0?x:len2:0;
return res;
}
}
#include <iostream>
#include <vector>
using namespace std;
int main(void)
{
int offset, currentpage, len1, len2;
while (cin >> offset >> currentpage >> len1 >> len2)
{
int start1 = 0, end1 = 0, start2 = 0, end2 = 0;
if (offset < len1)
{
start1 = offset;
if (currentpage > len1 - offset)
{
end1 = len1;
end2 = currentpage - (len1 - offset);
}
else
end1 = start1 + currentpage;
}
else
{
start1 = end1 = len1;
int temp = offset - len1;
if (temp > len2)
{
start2 = len2;
end2 = len2;
}
else
{
start2 = temp;
if (currentpage < len2 - start2)
end2 = start2 + currentpage;
else
end2 = len2;
}
}
cout << start1 << " " << end1 << " " << start2 << " " << end2 << endl;
}
return 0;
} 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
* @author wylu
*/
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
while ((line = br.readLine()) != null) {
String[] strs = line.split(" ");
int offset = Integer.parseInt(strs[0]), n = Integer.parseInt(strs[1]);
int len1 = Integer.parseInt(strs[2]), len2 = Integer.parseInt(strs[3]);
int s1 = Math.min(offset, len1);
int e1 = Math.min(offset + n, len1);
int s2 = Math.min(len2, Math.max(0, offset - len1));
int e2 = Math.max(0, Math.min(len2, offset + n - len1));
System.out.println(s1 + " " + e1 + " " + s2 + " " + e2);
}
}
}
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextInt()) { // 注意 while 处理多个 case
int offset = in.nextInt();
int n = in.nextInt();
int l1 = in.nextInt();
int l2 = in.nextInt();
// 根据offset计算起点
int l3 = l1 + l2;
// int plus = offset % l3;
int plus = offset;
// 判断偏移是否已经大于一遍
if (plus < l3) {
// 判断偏移是否超过l1
if (plus < l1) {
// 判断l1剩余空间是否够n
if (l1 - plus + 1 >= n) {
System.out.print(plus + " " + (plus + n - 1 + 1) + " " + "0 0" );
} else {
System.out.print(plus + " " + l1);
n -= (l1 - 1 - plus + 1);
// 判断l2剩余空间是否够n'
if (l2 >= n) {
System.out.print(" 0 " + (n - 1 + 1));
} else {
System.out.print(" 0 " + l2);
}
}
} else {
// 同理
System.out.print(l1 + " " + l1 + " ");
plus -= l1;
if (l2 - plus + 1 >= n) {
System.out.print(plus + " " + (plus + n - 1 + 1));
} else {
System.out.print(plus + " " + l2);
}
}
} else {
System.out.print(l1 + " " + l1 + " " + l2 + " " + l2);
}
System.out.println();
}
}
} 
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNextInt()){
int offset = sc.nextInt();
int n = sc.nextInt();
int l1 = sc.nextInt(), l2 = sc.nextInt();
int[] ans = solution(offset,n,l1,l2);
StringBuilder ss = new StringBuilder();
for(int i = 0; i < ans.length; i++){
if(i != ans.length - 1) ss.append(ans[i]).append(' ');
else ss.append(ans[i]);
}
System.out.println(ss.toString());
}
}
// 2 4 4 4 ==> 2 4 0 2
// 1 2 4 4 ==> 1 3 0 0
// 4 1 3 3 ==> 3 3 1 2
private static int[] solution(int offset, int n, int l1, int l2){
int[] ans = new int[4];
// 如果之前页面浏览的商品数量 offset 大于后面两列的数量，则表示 l1 和 l2 都需要表示浏览过的商品
// 如 4,1(可以是任意值),2,1
if(offset >= l1 + l2){
ans[0] = ans[1] = l1;
ans[2] = ans[3] = l2;
// 如果 offset >= l1 表示 l1 需要全部用来表示浏览过的，而 l2 用来表示全部或部分当前页展示的商品数量
// 如 3,2,2,2(l2 只能展示部分) 或 3,4,2,3(l2 展示全部)
} else if(offset >= l1){
ans[0] = ans[1] = l1;
ans[2] = offset - l1;
ans[3] = Math.min(l2,n + ans[2]);
// 如果 offset < l1 表示 l1 部分用来展示之前浏览的，剩下可以展示部分或全部
// 如 1,2,3,2(l1 展示全部) 或 1,3,2,3(l1 展示部分)
} else if(offset < l1){
ans[0] = offset;
ans[1] = Math.min(offset + n, l1);
ans[2] = 0;
ans[3] = offset + n <= l1 ? 0 : Math.min(offset + n - l1,l2);
}
return ans;
}
} 
import sys
def display(offset, n, num1, num2):
if offset <= num1:
start1 = offset
if n < num1 - offset:
end1 = n + offset
start2 = 0
end2 = 0
else:
end1 = num1
res = n + offset - num1
start2 = 0
end2 = min(res, num2)
else:
start1 = num1
end1 = num1
twoUsed = offset - num1
start2 = min(twoUsed, num2)
end2 = min(twoUsed + n, num2)
return start1, end1, start2, end2
for line in sys.stdin:
offset, n, num1, num2 = [int(i) for i in line.strip().split()]
start1, end1, start2, end2 = display(offset, n, num1, num2)
print(' '.join(map(str, [start1, end1, start2, end2])))
import java.util.Scanner;
public class ListCompletion {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int offset = sc.nextInt();//用户已经浏览的商品数
int n = sc.nextInt();//当前页面需要展示的商品数量
int l1 = sc.nextInt();//列表 L1 的长度
int l2 = sc.nextInt();//列表 L2 的长度
int start1 = 0;
int end1 = 0;
int start2 = 0;
int end2 = 0;
if (offset > l1 + l2) {
start1 = l1;
end1 = l1;
start2 = l2;
end2 = l2;
} else {
if (offset >= l1) {
start1 = l1;
end1 = l1;
start2 = (offset - l1);
if (offset + n >= l1 + l2) {
end2 = l2;
} else {
end2 = offset + n - l1;
}
} else {
start1 = offset;
if (offset + n <= l1) {
end1 = offset + n;
start2 = 0;
end2 = 0;
} else {
end1 = l1;
start2 = 0;
end2 = offset + n - l1;
}
}
}
System.out.println(start1 + " " + end1 + " " + start2 + " " + end2);
}
}
#include <bits/stdc++.h>
using namespace std;
int main (){
int offset,n,l1,l2;
int x,y,z,k;
while(cin>>offset>>n>>l1>>l2){
if(offset<=l1&&n>0){
x=(l1-offset==0?l1:offset);
y=min(l1,offset+n);
z=0;
if (l2>0)
k=max(0,offset+n-l1);
else
k=0;
}
if (offset>=l1&&n>0){
x=y=l1;
z=offset-l1;
k=min(offset-l1+n,l2);
}
if (n==0){
if(offset<=l1){
x=y=offset;
z=k=0;
}
else{
x=y=l1;
z=k=offset-l1;
}
}
if (offset>l1+l2){
x=y=l1;
z=k=l2;
}
cout<<x<<' '<<y<<' '<<z<<' '<<k;
}
} 
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