pat 1051 Pop Sequence 解法2

#include<cstdio>#include<cmath>using namespace std;int a[1005];int main(){int n,m,k,i,j;while(scanf("%d%d%d",&m,&n,&k)!=EOF){while(k--){int min = 100000,max = -1,f = 1;for(i = 1;i<=n;i++){scanf("%d",&a[i]);if(a[i]-i+1>m)f = 0;if(a[i]>max)min = max = a[i];else {if(a[i]>min)f = 0;else min = a[i];}}printf("%s\n",f?"YES":"NO");}}    return 0;}