codeforces297B

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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

3 3 3
2 2 2
1 1 3

YES

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

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

codeforces297B

标签：title   pac   determine   ict   先后   col   std   output   nbsp   

原文地址：https://www.cnblogs.com/gaojunonly1/p/10803551.html