Codeforces Round #418 (Div. 2)-A. An abandoned sentiment from past-思维
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http://codeforces.com/contest/814/problem/A
给定你两个数组,让第二个数组中的数可以替换掉第一个串中的0,问你是否可以使a串不是递增的,如果可以输出YES,否则输出NO
我的做法是暴力。只考虑最差的情况,把b数组排序后,从大到小填入a数组,然后判断最长递增子序列是不是a的长度。。
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <bits/stdc++.h>using namespace std;const int maxn=2000;int main(){ int a[maxn],m; int b[maxn],n; scanf("%d%d",&m,&n); for(int i=1;i<=m;i++){ scanf("%d",&a[i]); } for(int i=1;i<=n;i++){ scanf("%d",&b[i]); } sort(b+1,b+n+1); int j=n; for(int i=1;i<=m;i++){ if(a[i]==0){ a[i]=b[j--]; } } vector<int>q; for(int i=1;i<=m;i++){ int l=lower_bound(q.begin(),q.end(),a[i])-q.begin();//因为是非递减,所以用lower if(l==q.size()) q.push_back(a[i]); else q[l]=a[i]; } if(q.size()!=m) puts("YES"); else puts("NO"); /*cout<<q.size()<<endl; for(int i=0;i<q.size();i++){ cout<<q[i]<<endl; }*/ return 0;}
方法2就好多了。
当m大于1时肯定是yes的。。。(有很多种顺序)
当为1时判断就行,。。
#include<bits/stdc++.h> using namespace std; int main() { int a[201],b[201]; int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } for(int j=1;j<=m;j++) { scanf("%d",&b[j]); } if(m>1) { printf("Yes\n"); } if(m==1) { for(int i=1;i<=n;i++) { if(a[i]==0) a[i]=b[1]; } int t=1; for(int i=1;i<n;i++) { if(a[i]<a[i+1]) continue; else { t=0; break; } } if(t==1) printf("No\n"); else printf("Yes\n"); } return 0; }
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