链接:
http://www.bnuoj.com/bnuoj/problem_show.php?pid=26473
http://www.bnuoj.com/bnuoj/contest_show.php?cid=2317#problem/25675
Aaah!
Time Limit: 1000ms
Memory Limit: 65536KB
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Jon Marius shouted too much at the recent Justin Bieber concert, and now needs to go to the doctor because of his sore throat. The doctor’s instructions are to say “aaah”. Unfortunately, the doctors sometimes need Jon Marius to say “aaah” for a while, which Jon Marius has never been good at. Each doctor requires a certain level of “aah” – some require “aaaaaah”, while others can actually diagnose his throat with just a “h”. (They often diagnose wrongly, but that is beyond the scope of this problem.) Since Jon Marius does not want to go to a doctor and have his time wasted, he wants to compare how long he manages to hold the “aaah” with the doctor’s requirements. (After all, who wants to be all like “aaah” when the doctor wants you to go “aaaaaah”?)
Each day Jon Marius calls up a different doctor and asks them how long his “aaah” has to be. Find out if Jon Marius would waste his time going to the given doctor.
Input
The input consists of two lines. The first line is the “aaah” Jon Marius is able to say that day. The second line is the “aah” the doctor wants to hear. Only lowercase ’a’ and ’h’ will be used in the input, and each line will contain between 0 and 999 ’a’s, inclusive, followed by a single ’h’.
Output
Output “go” if Jon Marius can go to that doctor, and output “no” otherwise.
Sample Input
Sample Output
Source
The 2012 Nordic Collegiate Programming Contest
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题意:
第一个字符串给你 Jon 能发出的声音
第二个字符串给的是医生要求听到的声音,题目中已经说明了,医生必须听到他要听的声音才能诊断。
所以如果 Jon 如果不能发出医生要求的声音,去看医生就是浪费时间。
从而题目就变成了如果第一个字符串包含了第二个字符串,则输出 go
否则输出 no
算法:STL string 字符串查找
总结:
签到题很简单了,却不能快速看明白题意,秒杀掉,实在是弱爆了。
关于string 的find() 如果查到,则返回下标值(从 0 开始),如果查不到则返回 4294967295
刚刚去查了下这个数字
点击打开链接
4294967295就是无符号位的整形最大值,也就是有符号位整形-1,也就是s.find(" ") == -1,结果就显而易见了。
所以 string s1,s2;
如果 s1中不包含 s2
令int ans = s1.find(s2)
则 ans = -1
或者说 ans = 4294967295
code:
所以像下面这么写都可以水过。
#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
string s1,s2;
const int INF = 4294967295;
int main()
{
while(cin>>s1>>s2)
{
int ans = s1.find(s2);
if(ans != INF) printf("go\n");
else printf("no\n");
}
}
#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
string s1,s2;
const int INF = 4294967295; //表示找不到
int main()
{
while(cin>>s1>>s2)
{
int ans = s1.find(s2);
if(ans == -1) printf("no\n");
else printf("go\n");
}
}
#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
string s1,s2;
const int INF = 4294967295; //表示找不到
int main()
{
while(cin>>s1>>s2)
{
int ans = s1.find(s2);
if(ans >= 0)printf("go\n");
else printf("no\n");
}
}