Codeforces Round #429 (Div. 2) ABC

传送门

A. Generous Kefa

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

One day Kefa found n baloons. For convenience, we denote color of i-th
baloon as si —
lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa
want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES»,
if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 100)
— the number of baloons and friends.

Next line contains string s — colors of baloons.

Output

Answer to the task — «YES» or «NO» in a single line.

You can choose the case (lower or upper) for each letter arbitrary.

Examples

input

4 2
aabb

output

YES

input

6 3
aacaab

output

NO

Note

In the first sample Kefa can give 1-st and 3-rd
baloon to the first friend, and 2-nd and 4-th
to the second.

In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

题意：就是每个人拿的气球中不能是同色的，也就是看最多颜色那个气球与k比较大小。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

char s[maxn];
int a[maxn];
int main()
{
int n,k;
cin>>n>>k;
cin>>s+1;
int maxc=0;
FOR(1,n,i)
{
int x=s[i]-'a';
a[x]++;
maxc=max(x,maxc);
}
FOR(0,maxc,i)
{
if(a[i]>k)
{
puts("NO");
return 0;
}
}
puts("YES");
}

B. Godsend

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array.
Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player
can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

Input

First line of input data contains single integer n (1 ≤ n ≤ 106)
— length of the array.

Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Output answer in single line. "First", if first player wins, and "Second"
otherwise (without quotes).

Examples

input

4
1 3 2 3

output

First

input

2
2 2

output

Second

Note

In first sample first player remove whole array in one move and win.

In second sample first player can't make a move and lose.

题意：一个人可以拿区间和加起来是奇数的区间，然后从总区间中去掉。另外一个拿可以拿偶数的区间，然后去掉。

也就是看有没有奇数出现，如果有奇数出现就是First。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int n;
LL x;
int main()
{

cin>>n;
for(int i=1;i<=n;i++)
{
scan_d(x);
if(x&1)
{
puts("First");
return 0;
}
}
puts("Second");
}

C. Leha and Function

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Leha like all kinds of strange things. Recently he liked the function F(n, k). Consider all possible k-element
subsets of the set [1, 2, ..., n]. For subset find minimal element in it. F(n, k) —
mathematical expectation of the minimal element among all k-element subsets.

But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays A and B,
each consists of m integers. For all i, j such
that 1 ≤ i, j ≤ m the condition Ai ≥ Bj holds.
Help Leha rearrange the numbers in the array A so that the sum 

 is
maximally possible, where A' is already rearranged array.

Input

First line of input data contains single integer m (1 ≤ m ≤ 2·105)
— length of arrays A and B.

Next line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 109)
— array A.

Next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109)
— array B.

Output

Output m integers a'1, a'2, ..., a'm —
array A' which is permutation of the array A.

Examples

input

5
7 3 5 3 4
2 1 3 2 3

output

4 7 3 5 3

input

7
4 6 5 8 8 2 6
2 1 2 2 1 1 2

output

2 6 4 5 8 8 6

垃圾题意，读了一个小时。

题意：让每一个n+1/k+1最大。

比如样例1.

7 3 5 3 4 

2 1 3 2 3

让7/1才会让n+1/k+1最大，也就是说让n大,k小。

所以

4 7 3 5 3
2 1 3 2 3

对应的才是使n+1/k+1最大的，因为B序列的位置不变，所以只能对A变位置。

先排两次序，找到一一对应的A B，变回B原先的位置就好了。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int m,a[maxx],b[maxx];
struct node
{
int pos, w;
}Q[maxx];

bool cmp(node a,node b)
{
return a.w>b.w;
}
int main()
{
scan_d(m);
for(int i=1;i<=m;i++)
{
scan_d(a[i]);
}
int x;
sort(a+1,a+m+1);
for(int i=1;i<=m;i++)
{
scan_d(Q[i].w);
Q[i].pos=i;
}
sort(Q+1,Q+m+1,cmp);
FOR(1,m,i)
{
Q[i].w=a[i];
}
FOR(1,m,i)
{
a[Q[i].pos]=Q[i].w;
}
FOR(1,m,i)
{
cout<<a[i]<<" ";
}
cout<<endl;
}