题目链接：https://vjudge.net/contest/216713
两个月前，也就是寒假做的题。
据说是用来熟悉STL的水题，然而不熟练STL的我做得好艰难啊。
只按照书上的最低要求选做了一些，其他实在是懒得做…
例题5-1
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
const int maxn = 10000 + 10;
int a[maxn];
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n, q, kase = 0;
while(scanf("%d%d", &n, &q) && n && q)
{
printf("CASE# %d:\n", ++kase);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
sort(a, a + n);
int x, flag;
for(int i = 0; i < q; i++)
{
scanf("%d", &x);
flag = 0;
for(int j = 0; j < n; j++)
{
if(a[j] == x)
{
flag = 1;
printf("%d found at %d\n", x, j + 1);
break;
}
}
if(!flag)
printf("%d not found\n", x);
}
}
return 0;
}
例题5-2
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
vector<int> v[]相当于二维数组，
它的第一维大小固定，第二维大小不固定；
题目有四个指令，善于寻找他们的共同点
*/
vector<int> v[30];
int n;
//寻找x木块所在堆和高度，用引用返回
void Find(int x, int &p, int &h)
{
for(p = 0; p < n; p++)
for(h = 0; h < v[p].size(); h++)
if(v[p][h] == x)
return;
}
//把x木块上面的木块归位,resize
void POP(int x, int p, int h)
{
for(int i = h + 1; i < v[p].size(); i++)
{
int x = v[p][i];
v[x].push_back(x);
}
v[p].resize(h + 1);
}
//把a及上方木块移动到b上方
void Move(int a, int b, int pa, int ha, int pb, int hb)
{
for(int i = ha; i < v[pa].size(); i++)
{
int x = v[pa][i];
v[pb].push_back(x);
}
v[pa].resize(ha);
}
void Print()
{
for(int i = 0; i < n; i++)
{
cout << i << ':';
for(int j = 0; j < v[i].size(); j++)
cout << ' ' << v[i][j];
cout << '\n';
}
}
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
cin >> n;
for(int i = 0; i < n; i++)
v[i].push_back(i);
string s1, s2;
int a, b;
while(cin >> s1 >> a >> s2 >> b)
{
int pa, ha, pb, hb;
Find(a, pa, ha);
Find(b, pb, hb);
if(pa == pb || a == b)
continue;
if(s2 == "onto")
POP(b, pb, hb);
if(s1 == "move")
POP(a, pa, ha);
Move(a, b, pa, ha, pb, hb);
}
Print();
return 0;
}
例题5-3
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
学会使用stringstream，迭代器iterator和set；
set中各个元素不重复且有序
*/
set<string> S;
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
string s, a;
while(cin >> s)
{
for(int i = 0; i < s.length(); i++)
{
if(isalpha(s[i]))
{
if(s[i] >= 'A' && s[i] <= 'Z')
s[i] = s[i] + 32;
// s[i] = tolower(s[i]);
}
else
s[i] = ' ';
}
stringstream ss(s);
while(ss >> a)
{
S.insert(a);
}
}
set<string>::iterator it;
for(it = S.begin(); it != S.end(); it++)
cout << *it << '\n';
return 0;
}
例题5-4
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
开两个vector存原始单词和转换成小写后的单词；
map可以计数
*/
vector<string> words;
vector<string> ans;
map<string, int> num;
string tra(string s)
{
string ans = s;
for(int i = 0; i < ans.length(); i++)
ans[i] = tolower(ans[i]);
sort(ans.begin(), ans.end());
return ans;
}
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
string s, a;
while(cin >> s && s != "#")
{
words.push_back(s);
a = tra(s);
if(!num.count(a))
num[a] = 0;
num[a]++;
}
for(int i = 0; i < words.size(); i++)
{
if(num[tra(words[i])] == 1)
ans.push_back(words[i]);
}
sort(ans.begin(), ans.end());
for(int i = 0; i < ans.size(); i++)
cout << ans[i] << '\n';
return 0;
}
例题5-5
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
关键:
set<int> Set;
map<Set, int> ID;
vector<Set> SET;
对于每一个Set s,
ID[s]是ID，SET[ID[s]]是s本身。
*/
typedef set<int> Set;
map<Set, int> ID;
vector<Set> SET;
int Search(Set s)
{
if(!ID.count(s))
{
SET.push_back(s);
ID[s] = SET.size() - 1;
}
return ID[s];
}
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int T, n;
scanf("%d", &T);
while(T--)
{
stack<int> s;
scanf("%d", &n);
while(n--)
{
string p;
cin >> p;
if(p[0] == 'P')
s.push(Search(Set()));
else if(p[0] == 'D')
s.push(s.top());
else
{
Set x1 = SET[s.top()];
s.pop();
Set x2 = SET[s.top()];
s.pop();
Set x;
if(p[0] == 'U')
set_union(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x, x.begin()));
else if(p[0] == 'I')
set_intersection(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x, x.begin()));
else
{
x = x2;
x.insert(Search(x1));
}
s.push(Search(x));
}
cout << SET[s.top()].size() << '\n';
}
cout << "***" << '\n';
}
return 0;
}
例题5-6
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
要弄清题意
*/
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int T, kase = 1;
while(scanf("%d", &T) && T)
{
queue<int> Q;
queue<int> q[1010];
map<int, int> M;
for(int i = 0; i < T; i++)
{
int n;
scanf("%d", &n);
while(n--)
{
int x;
scanf("%d", &x);
M[x] = i;
}
}
printf("Scenario #%d\n", kase++);
char s[10];
while(1)
{
scanf("%s", s);
if(s[0] == 'S')
break;
else if(s[0] == 'E')
{
int x;
scanf("%d", &x);
int t = M[x];
if(q[t].empty())
Q.push(t);
q[t].push(x);
}
else
{
int t = Q.front();
printf("%d\n", q[t].front());
q[t].pop();
if(q[t].empty())
Q.pop();
}
}
printf("\n");
}
return 0;
}
例题5-7
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
priority_queue<ll, vector<ll>, greater<ll> > pq;
优先队列，越大优先级越小，即从小到大排序。
若省略最后一个参数，则是从大到小。
*/
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
priority_queue<ll, vector<ll>, greater<ll> > pq;
set<ll> S; //用于判断是否重复
pq.push(1);
S.insert(1);
for(int i = 1; ; i++)
{
ll x = pq.top();
pq.pop();
if(i == 1500)
{
printf("The 1500'th ugly number is %lld.\n", x);
break;
}
ll y;
y = 2 * x;
if(!S.count(y))
{
pq.push(y);
S.insert(y);
}
y = 3 * x;
if(!S.count(y))
{
pq.push(y);
S.insert(y);
}
y = 5 * x;
if(!S.count(y))
{
pq.push(y);
S.insert(y);
}
}
return 0;
}
例题5-8
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
计算行列，模拟即可
*/
string s[110];
const int maxn = 60;
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n;
while(cin >> n)
{
int M = 0;
for(int i = 0; i < n; i++)
{
cin >> s[i];
M = max(M, (int)s[i].length());
}
sort(s, s + n);
int cols = (60 - M) / (M + 2) + 1;
int rows = (n - 1) / cols + 1;
for(int i = 0; i < maxn; i++)
cout << '-';
cout << '\n';
for(int i = 0; i < rows; i++)
{
for(int j = 0; j < cols; j++)
{
int x = rows * j + i;
if(x < n)
{
int len = s[x].length();
cout << s[x];
if(j == cols - 1)
{
for(int k = 0; k < M - len; k++)
cout << ' ';
}
else
{
for(int k = 0; k < M + 2 - len; k++)
cout << ' ';
}
}
}
cout << '\n';
}
}
return 0;
}
例题5-9
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
直接枚举r1,r2,c1,c2肯定T
解决方法：
枚举c1,c2,然后从上到下扫描各行
把每行c1,c2列的内容存到map中
若已经存在，则找到
*/
struct Node
{
int x, y;
Node(int a, int b)
{
x = a;
y = b;
}
bool operator < (const Node & b) const
{
if(x == b.x)
return y < b.y;
return x < b.x;
}
};
map<string, int> ID; //给每个字符串一个编号
vector<string> str; //保存编号对应的字符串
vector<int> v[10010]; //存处理好的int表格
map<Node, int> M;
int IDD(string s) //获取编号
{
if(ID.count(s))
return ID[s];
str.push_back(s);
return ID[s] = str.size() - 1;
}
void solve(int n, int m)
{
for(int i = 0; i < m; i++) //i列
{
for(int j = i + 1; j < m; j++) //j列
{
M.clear();
for(int k = 0; k < n; k++) //k行
{
Node node(v[k][i], v[k][j]);
if(M.count(node))
{
printf("NO\n");
printf("%d %d\n", M[node] + 1, k + 1);
printf("%d %d\n", i + 1, j + 1);
return;
}
else
{
M[node] = k;
}
}
}
}
printf("YES\n");
}
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n, m;
while(~scanf("%d%d", &n, &m))
{
string s;
char ch = getchar();
for(int i = 0; i < n; i++)
{
while(1)
{
ch = getchar();
if(ch == '\n' || ch == '\r')
{
if(!s.empty())
v[i].push_back(IDD(s));
s.clear();
break;
}
if(ch == ',')
{
v[i].push_back(IDD(s));
s.clear();
}
else
{
s = s + ch;
}
}
}
solve(n, m);
//千万不能忘！！！
for(int i = 0; i < n; i++)
v[i].clear();
ID.clear();
str.clear();
}
return 0;
}
习题5-1
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
用getline读一行
用stringstream一个个读入
注意getline格式
*/
vector<string> v[1010];
int a[200];
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int i = 0, num = 0;
string sin;
while(getline(cin, sin))
{
stringstream ss(sin);
string s;
while(ss >> s)
v[i].push_back(s);
i++;
}
//求第k个单词的最大长度存入a[]
for(int j = 0; j < i; j++)
{
for(int k = 0; k < v[j].size(); k++)
{
a[k] = max(a[k], (int)v[j][k].length());
num = max(num, k);
}
}
//输出，注意最后一列不用输出空格
for(int j = 0; j < i; j++)
{
for(int k = 0; k < v[j].size() - 1; k++)
{
cout << v[j][k];
for(int l = 0; l < a[k] - v[j][k].length(); l++)
cout << ' ';
cout << ' ';
}
cout << v[j][v[j].size() - 1] << '\n';
}
return 0;
}
习题5-2
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
题目保证1000步之内必定变成0或者循环，所以判断0即可
*/
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int a[2][20];
int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%d", &a[0][i]);
}
int j = 0;
bool flag = false;
for(int l = 0; l < 2000; l++)
{
bool zero = true;
int k = j ? 0 : 1;
for(int i = 0; i < n; i++)
{
int m = (i + 1) % n;
a[k][i] = fabs(a[j][i] - a[j][m]);
if(a[k][i] != 0)
zero = false;
}
j = k;
if(zero)
{
printf("ZERO\n");
flag = true;
break;
}
}
if(!flag)
printf("LOOP\n");
}
return 0;
}
习题5-3
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
简单的队列操作
*/
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n;
while(~scanf("%d", &n) && n)
{
bool first = true;
queue<int> q;
for(int i = 1; i <= n; i++)
q.push(i);
printf("Discarded cards:");
while(1)
{
if(q.size() == 1)
{
printf("\nRemaining card: %d\n", q.front());
break;
}
int x = q.front();
if(first)
{
first = false;
printf(" %d", x);
}
else
printf(", %d", x);
q.pop();
x = q.front();
q.pop();
q.push(x);
}
}
return 0;
}
习题5-4
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
要读懂题意！！！
题目意思是，申请a到b和b到a的数量一样就OK。
用数组，每输入两个数就交换
如果最后数组和原数组相同则OK
a,b b,c c,a的情况也行
*/
const int maxn = 500000 + 10;
int arr[maxn];
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n;
while(~scanf("%d", &n) && n)
{
for(int i = 0; i < n; i++)
arr[i] = i;
for(int i = 0; i < n; i++)
{
int a, b;
scanf("%d%d", &a, &b);
swap(arr[a], arr[b]);
}
bool flag = false;
for(int i = 0; i < n; i++)
{
if(arr[i] != i)
{
flag = true;
printf("NO\n");
break;
}
}
if(!flag)
printf("YES\n");
}
return 0;
}
习题5-5
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
枚举每一个单词
如果再枚举两个单词是否能拼成它，会超时
所以用substr把这个单词拆分
用map映射，在里面的单词映射成1，不在的0
*/
const int maxn = 120000 + 10;
map<string, bool> m;
string s[maxn];
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int n = 0;
while(cin >> s[n])
{
m[s[n++]] = 1;
}
string a, b;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < s[i].size() - 1; j++)
{
a = s[i].substr(0, j + 1);
if(!m[a])
continue;
b = s[i].substr(j + 1);
if(!m[b])
continue;
cout << s[i] << endl;
break;
}
}
return 0;
}
习题5-6
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
做了很长时间，题目要求考虑不周，代码能力，还忘记初始化
*/
struct node
{
int x, y;
bool flag; //判断是否找到了对称点
bool operator < (const node &b) const
{
if(x == b.x)
return y < b.y;
return x < b.x;
}
}node[1010];
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%d%d", &node[i].x, &node[i].y);
node[i].x *= 2; //为了避免出现小数，横坐标都乘2
node[i].flag = false; //初始化！！！
}
sort(node, node + n);
int mid = (node[0].x + node[n - 1].x) / 2;
bool ok = true;
for(int i = 0; i < n; i++)
{
if(node[i].x == mid) //在对称轴上
{
node[i].flag = true;
continue;
}
if(node[i].flag) //这点已经有对称点了
continue;
if(!ok) //存在点没有对称轴
break;
int a = 2 * mid - node[i].x;
for(int j = n - 1; j > 0; j--)
{
if(i == j) //同一点
continue;
if(node[j].flag) //这点已经有对称点了
continue;
if(node[j].x == a)
{
if(node[j].y == node[i].y)
{
node[j].flag = true;
node[i].flag = true;
break;
}
else
continue;
}
else if(node[j].x < a)
{
ok = false;
break;
}
else
continue;
}
}
bool yes = true;
if(!ok)
{
printf("NO\n");
yes = false;
}
else
{
for(int i = 0; i < n; i++)
{
if(!node[i].flag)
{
printf("NO\n");
yes = false;
break;
}
}
if(yes)
printf("YES\n");
}
}
return 0;
}
习题5-7
方法一：直接用数组模拟一个队列
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
模拟一个队列即可
*/
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int T;
scanf("%d", &T);
while(T--)
{
int q[150];
int front = 0, rear = 0;
int n, m, ans = 0;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
{
int x;
scanf("%d", &x);
q[rear++] = x;
}
while(1)
{
int x = q[front];
bool flag = false;
int i;
for(i = (front + 1) % 150; i != rear; i = (i + 1) % 150)
{
if(q[i] > x)
{
flag = true;
break;
}
}
if(flag)
{
if(m == front)
m = rear;
front = (front + 1) % 150;
q[rear] = x;
rear = (rear + 1) % 150;
}
else
{
ans++;
if(m == front)
{
printf("%d\n", ans);
break;
}
front = (front + 1) % 150;
}
}
}
return 0;
}
方法二：用优先队列
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<cstdlib>
#include<ctime>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
/*
可以用优先队列存权值
*/
struct node
{
int p; //权值
bool t; //是否是要求的
};
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int T;
scanf("%d", &T);
while(T--)
{
int n, m, ans = 0;
queue<node> q;
priority_queue<int> pq;
//普通的优先队列，从大到小排列,也可用数组存
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
{
node x;
scanf("%d", &x.p);
if(i == m)
x.t = true;
else
x.t = false;
q.push(x);
pq.push(x.p);
}
while(1)
{
node x = q.front();
int j = pq.top();
q.pop();
if(x.p == j)
{
ans++;
pq.pop();
if(x.t)
{
printf("%d\n", ans);
break;
}
}
else
{
q.push(x);
}
}
}
return 0;
}