We Love MOE Girls
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 205 Accepted Submission(s): 154
Problem Description
Chikami Nanako is a girl living in many different parallel worlds. In this problem we talk about one of them.
In this world, Nanako has a special habit. When talking with others, she always ends each sentence with "nanodesu".
There are two situations:
If a sentence ends with "desu", she changes "desu" into "nanodesu", e.g. for "iloveyoudesu", she will say "iloveyounanodesu". Otherwise, she just add "nanodesu" to the end of the original sentence.
Given an original sentence, what will it sound like aften spoken by Nanako?
In this world, Nanako has a special habit. When talking with others, she always ends each sentence with "nanodesu".
There are two situations:
If a sentence ends with "desu", she changes "desu" into "nanodesu", e.g. for "iloveyoudesu", she will say "iloveyounanodesu". Otherwise, she just add "nanodesu" to the end of the original sentence.
Given an original sentence, what will it sound like aften spoken by Nanako?
Input
The first line has a number T (T <= 1000) , indicating the number of test cases.
For each test case, the only line contains a string s, which is the original sentence.
The length of sentence s will not exceed 100, and the sentence contains lowercase letters from a to z only.
For each test case, the only line contains a string s, which is the original sentence.
The length of sentence s will not exceed 100, and the sentence contains lowercase letters from a to z only.
Output
For every case, you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output which Nanako will say.
Sample Input
2 ohayougozaimasu daijyoubudesu
Sample Output
Case #1: ohayougozaimasunanodesu Case #2: daijyoubunanodesu
Source
题目地址：We Love MOE Girls
AC代码：
#include<iostream>
#include<cstring>
#include<string>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
char p[105];
char t[5];
int main()
{
int tes,i;
int cas=0;
int len;
scanf("%d",&tes);
while(tes--)
{
scanf("%s",p);
len=strlen(p);
printf("Case #%d: ",++cas);
if(len<4)
{
cout<<p<<"nanodesu"<<endl;
}
else
{
t[0]=p[len-4];
t[1]=p[len-3];
t[2]=p[len-2];
t[3]=p[len-1];
t[4]='\0';
if(strcmp(t,"desu")==0)
{
for(i=0;i<len-4;i++)
cout<<p[i];
cout<<"nanodesu"<<endl;
}
else
cout<<p<<"nanodesu"<<endl;
}
}
return 0;
}
A Bit Fun
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 395 Accepted Submission(s): 254
Problem Description
There are n numbers in a array, as a0, a1 ... , an-1, and another number m. We define a function f(i, j) = ai|ai+1|ai+2| ... | aj . Where "|" is the bit-OR operation. (i <= j)
The problem is really simple: please count the number of different pairs of (i, j) where f(i, j) < m.
The problem is really simple: please count the number of different pairs of (i, j) where f(i, j) < m.
Input
The first line has a number T (T <= 50) , indicating the number of test cases.
For each test case, first line contains two numbers n and m.(1 <= n <= 100000, 1 <= m <= 230) Then n numbers come in the second line which is the array a, where 1 <= ai <= 230.
For each test case, first line contains two numbers n and m.(1 <= n <= 100000, 1 <= m <= 230) Then n numbers come in the second line which is the array a, where 1 <= ai <= 230.
Output
For every case, you should output "Case #t: " at first, without quotes. The t is the case number starting from 1.
Then follows the answer.
Then follows the answer.
Sample Input
2 3 6 1 3 5 2 4 5 4
Sample Output
Case #1: 4 Case #2: 0
Source
题目大意：题目意思很简单，只是当时真的不敢暴力啊，(10^5)^2。直接暴力写出来了。。
题目地址：A Bit Fun
AC代码：
#include<iostream>
#include<cstring>
#include<string>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int a[100005];
int main()
{
int tes,i,j;
int cas=0;
scanf("%d",&tes);
while(tes--)
{
int n,m;
int res=0;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
int tmp=0; //0|p=p
for(j=i;j<n;j++)
{
tmp=tmp|a[j];
if(tmp>=m)
break;
res++;
}
}
printf("Case #%d: %d\n",++cas,res);
}
return 0;
}
//171MS 620K