Codeforces Round #271 (Div. 2)

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原创最新推荐文章于 2026-06-21 14:59:16 发布 1.1k 阅读 AI 写同款· GEO 优化 ›

基于libVLC的视频播放器之四：直接使用libVLC 一.效果播放本地文件播放网络url：rtsp://3.84.6.190/vod/mp4:BigBuckBunny_115k.mov 二.实现既然有VLC-Qt，为什么还要直接调用libVLC呢，因为直接调用会加深对整个播放流程的理解，方便后面修改VLC-Qt源码。注意事项： 1.编译报错：error C2065: “libvlc_media_read_cb”: 未声明的标识符。修改如下： 2.libvlc_new(0, nullptr)返回null 将plugins 阅读详情

Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:

qwertyuiop
asdfghjkl;
zxcvbnm,./

Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).

We have a sequence of characters he has typed and we want to find the original message.

Input

First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).

Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.

It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.

Output

Print a line that contains the original message.

        Sample test(s)
      
          input
        
R
s;;upimrrfod;pbr

          output
        
allyouneedislove

直接模拟即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int INF=100000000;
const int N=110;
char s[N],ss[N];
int main()
{
char ch;
char str[N];
strcpy(str,"qwertyuiopasdfghjkl;zxcvbnm,./");
while(scanf("%c",&ch)!=EOF)
{
getchar();
gets(s);
int len=strlen(s);
//puts(str);
if(ch=='L')
{
for(int i=0;i<len;i++)
{
int cnt=strchr(str,s[i])-str;
ss[i]=str[cnt+1];
}
}
else
{
//cout<<"in"<<endl;
for(int i=0;i<len;i++)
{
int cnt=strchr(str,s[i])-str;
//cout<<"cnt="<<cnt<<endl;
ss[i]=str[cnt-1];
}
}
ss[len]='\0';
puts(ss);

}
return 0;
}

It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.

Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.

Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.

Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.

Input

The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.

The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.

The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.

Output

Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.

        Sample test(s)
      

          input
        
5
2 7 3 4 9
3
1 25 11

          output
        
1
5
3

Note

For the sample input:

The worms with labels from [1, 2] are in the first pile.
The worms with labels from [3, 9] are in the second pile.
The worms with labels from [10, 12] are in the third pile.
The worms with labels from [13, 16] are in the fourth pile.
The worms with labels from [17, 25] are in the fifth pile.

sum[i] 为前 i 项的和，lower_bound判断一下就行

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int INF=100000000;
const int N=100010;
int a[N],sum[N];
int main()
{
int n,q;
while(~scanf("%d",&n))
{
sum[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
scanf("%d",&q);
while(q--)
{
int t;
scanf("%d",&t);
int cnt=lower_bound(sum+1,sum+1+n,t)-sum;
printf("%d\n",cnt);
}
}
return 0;
}