Strange Class

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

In Vivid’s school, there is a strange class(SC). In SC, the students’ names are very strange. They are in the same format: anbncn(a,b,c
must not be the same with each other). For example studens whose names are“abc”,”ddppqq” are in SC, however studens whose names are “aaa”,“ab”,”ddppqqq” are not in SC.

Vivid makes friends with so many students, he wants to know who are in SC.

Input

There are multiple test cases (about 10), each case will give a string S which is the name of Vivid’s friend in a single line.

Please process to the end of file.

[Technical Specification]

1≤|S|≤10.

|S| indicates the length of S.

S only contains lowercase letter.

Output

For each case, output YES if Vivid’s friend is the student of SC, otherwise output NO.

Sample Input

abc
bc

Sample Output

YES
NO

Source

BestCoder Round #36 ($)

很猥琐的一道题

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

char str[350];

int main() {
    while(scanf("%s", str) != EOF) {
        int len = strlen(str);
        if(len <= 2 || len % 3 != 0 || str[0] == str[len - 1]) {
            printf("NO\n");
            continue;
        }

        int a[5] = {0, 0, 0};
        int sum = 0;
        char cur = str[0];
        for(int i = 0; i < len; i++) {
            if(str[i] == cur) a[sum]++;
            else {
                sum++;
                a[sum] ++;
                cur = str[i];
            }
        }

        //printf("%d %d %d %d\n", a[0], a[1], a[2], sum);
        int tmp = len / 3;
        if((len % 3 == 0) && sum == 2 && a[0] == tmp && a[1] == tmp && a[2] == tmp) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Long long ago, there is a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds
and n trees.
The i?th bird
stands on the top of the i?th tree.
The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the bird which stands in the tree with height H will
falls.

Jack will shot many times, he wants to know how many birds fall during each shot.

a bullet can hit many birds, as long as they stand on the top of the tree with height of H.

Input

There are multiple test cases (about 5), every case gives n,m in
the first line, n indicates
there are n trees
and n birds, m means
Jack will shot m times.

In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which
describes the height of the trees.

In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which
describes the height of the Jack’s shots.

Please process to the end of file.

[Technical Specification]

1≤n,m≤1000000(106)

1≤h[i],q[i]≤1000000000(109)

All inputs are integers.

Output

For each q[i],
output an integer in a single line indicates the number of birds Jack shot down.

Sample Input

4 3
1 2 3 4
1 1 4

Sample Output

1
0
1

Hint

Huge input, fast IO is recommended.

Source

BestCoder Round #36 ($)

卡输入，要快速读入，不过貌似不写快速读入也行╮(╯▽╰)╭，心累啊，我还多交了一遍

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <map>
using namespace std;

int n, m;

inline int readint()
{
    char c = getchar();
    while(!isdigit(c)) c = getchar();  

    int x = 0;
    while(isdigit(c))
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}  

int buf[10];
inline void writeint(int i)
{
    int p = 0;
    if(i == 0) {
        p++;
        buf[0] = 0;
    }
    else while(i)
    {
        buf[p++] = i % 10;
        i /= 10;
    }
    for(int j = p-1; j >= 0; j--) putchar('0' + buf[j]);
}  

int main() {
    while(scanf("%d %d", &n, &m) != EOF) {
        map<int, int> mp;
        for(int i = 0; i < n; i++) {
            int t = readint();
            mp[t] ++;
        }

        for(int i = 0; i < m; i++) {
            int t = readint();
            //printf("%d\n", mp[t]);
            writeint(mp[t]);
            putchar(10);
            mp[t] = 0;
        }
    }
    return 0;
}