Codeforces Round #180 (Div. 1) B

B. Fish Weight
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 wi, then 0 < w1 ≤ w2 ≤ … ≤ wk 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 wi (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 = 6 weight 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.

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;int a[123456],b[123456];int k;int main(){    int n,m,k;    scanf("%d %d %d",&n,&m,&k);    for(int i=0;i<n;i++) scanf("%d",&a[i]);    for(int i=0;i<m;i++) scanf("%d",&b[i]);    if(n<m) for(int i=n;i<m;i++) a[i]=0;    if(m<n) for(int i=m;i<n;i++) b[i]=0;    sort(a,a+max(n,m));    sort(b,b+max(n,m));    for(int i=0;i<max(n,m);i++) if(a[i] > b[i]) return 0*puts("YES");    puts("NO");    return 0;}