codeforces297B

最新推荐文章于 2025-05-29 09:09:56 发布

转载最新推荐文章于 2025-05-29 09:09:56 发布 · 191 阅读

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原文链接：http://www.cnblogs.com/gaojunonly1/p/10803551.html

The blog discusses a fish weight problem from CodeForces. Given the number of fish caught by Alice and Bob, the number of fish species, and the types of fish they caught, it aims to determine if Alice's fish can have a strictly larger total weight. It provides input and output requirements, examples, and a solution idea of using greedy algorithm.

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Fish Weight

CodeForces - 297B

It is known that there are k fish species in the polar ocean, numbered from 1 to k. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the i-th type of fish be w_i, then 0 < w₁ ≤ w₂ ≤ ... ≤ w_k holds.

Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights w_i (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 10⁵, 1 ≤ k ≤ 10⁹) — the number of fish caught by Alice and Bob respectively, and the number of fish species.

The second line contains n integers each from 1 to k, the list of fish type caught by Alice. The third line contains m integers each from 1 to k, the list of fish type caught by Bob.

Note that one may have caught more than one fish for a same species.

Output

Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.

Examples

Input

3 3 3
2 2 2
1 1 3

Output

YES

Input

4 7 9
5 2 7 3
3 5 2 7 3 8 7

Output

NO

Note

In the first sample, if w₁ = 1, w₂ = 2, w₃ = 2.5, then Alice has a total of 2 + 2 + 2 = 6weight units, while Bob only has 1 + 1 + 2.5 = 4.5.

In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.

sol：贪心乱搞，先后后缀和只要个数大于另一个就是YES了

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=100005;
int n,m,K,a[N],b[N],Hash[N<<1];
int main()
{
    int i;
    R(n); R(m); R(K);
    for(i=1;i<=n;i++) Hash[++*Hash]=(a[i]=read());
    for(i=1;i<=m;i++) Hash[++*Hash]=(b[i]=read());
    sort(Hash+1,Hash+*Hash+1);
    *Hash=unique(Hash+1,Hash+*Hash+1)-Hash-1;
    for(i=1;i<=n;i++) a[i]=lower_bound(Hash+1,Hash+*Hash+1,a[i])-Hash;
    for(i=1;i<=m;i++) b[i]=lower_bound(Hash+1,Hash+*Hash+1,b[i])-Hash;
    sort(a+1,a+n+1);
    sort(b+1,b+m+1);
    int Now=m+1;
    for(i=n;i>=1;i--)
    {
        while(Now>1&&b[Now-1]>=a[i]) Now--;
        if(n-i+1>m-Now+1) return puts("YES"),0;
    }
    puts("NO");
    return 0;
}
/*
Input
3 3 3
2 2 2
1 1 3
Output
YES

Input
4 7 9
5 2 7 3
3 5 2 7 3 8 7
Output
NO
*/

View Code

转载于:https://www.cnblogs.com/gaojunonly1/p/10803551.html