HDU 2141 Can you find it?
Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
Output
For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.
Sample Input
3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO
题意
给出三个数组，每个数组选一个数相加，让其等于下面给出的和
二分查找
将a + b + c = s 视为 s - a = b + c
将b + c形成一个新数组k，并sort（排序，二分查找的前提）
一个for循环(s - a)
暴力超时
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int L, N, M;
int l[505];
int n[505];
int m[505];
int k[250005];
bool diy_search(int x)
{
int l = 0, r = M * N;
while(r - l >= 1)
{
int i = (l + r) / 2;
if(k[i] == x)
return 1;
else if(k[i] < x)
l = i + 1;
else
r = i;
}
return 0;
}
int main()
{
int tot = 1;
while(~scanf("%d%d%d", &L, &N, &M))
{
printf("Case %d:\n", tot);
tot++;
for(int i = 0; i < L; ++i)
scanf("%d", &l[i]);
for(int i = 0; i < N; ++i)
scanf("%d", &n[i]);
for(int i = 0; i < M; ++i)
scanf("%d", &m[i]);
for(int i = 0; i < N; ++i)
for(int j = 0; j < M; ++j)
k[i * N + j] = n[i] + m[j];
sort(k, k + N * M);
int s;
scanf("%d", &s);
for(int i = 0; i < s; ++i)
{
int tem;
scanf("%d", &tem);
bool flag = 0;
for(int j = 0; j < L; ++j)
if(diy_search(tem - l[j]))
{
flag = 1;
cout << "YES" << '\n';
goto over;///goto跳出多重循环
}
if(!flag)
cout << "NO" << '\n';
over:
;
}
}
return 0;
}
用(upper_bound - lower_bound > 0)代替自己写的二分查找函数
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int L, N, M;
int l[505];
int n[505];
int m[505];
int k[250005];
int main()
{
int tot = 1;
while(~scanf("%d%d%d", &L, &N, &M))
{
printf("Case %d:\n", tot);
tot++;
for(int i = 0; i < L; ++i)
scanf("%d", &l[i]);
for(int i = 0; i < N; ++i)
scanf("%d", &n[i]);
for(int i = 0; i < M; ++i)
scanf("%d", &m[i]);
for(int i = 0; i < N; ++i)
for(int j = 0; j < M; ++j)
k[i * N + j] = n[i] + m[j];
sort(k, k + N * M);
int s;
scanf("%d", &s);
for(int i = 0; i < s; ++i)
{
int tem;
scanf("%d", &tem);
bool flag = 0;
for(int j = 0; j < L; ++j)
if(upper_bound(k, k + N * M, tem - l[j]) - lower_bound(k, k + N * M, tem - l[j]))
{
flag = 1;
cout << "YES" << '\n';
goto over;
}
if(!flag)
cout << "NO" << '\n';
over:
;
}
}
return 0;
}
关于goto
goto 语句可以无条件地转移到过程中指定的行。
goto 语句通常与条件语句配合使用。可用来实现条件转移， 构成循环，跳出循环体等功能。
弊端
在结构化程序设计中一般不主张使用 goto 语句， 以免造成程序流程的混乱，使理解和调试程序都产生困难。