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2018暑期多校1

分类: IT文章 • 2024-08-17 20:03:07

1001

Problem Description

Given an integer z is maximum.
 

Input

There are multiple test cases. The first line of input contains an integer 6).

 Output

For each test case, output an integer denoting the maximum 1 instead.

 Sample Input

3
1
2
3

 Sample Output

-1
-1
1
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;
 6     scanf("%d",&T);
 7     while (T--)
 8     {
 9         long long n;
10         scanf("%lld",&n);
11         if(n%3==0)
12         {
13             printf("%lld
",(n/3)*(n/3)*(n/3));
14         } else if(n%4==0)
15         {
16             printf("%lld
",(n/2)*(n/4)*(n/4));
17         }
18         else printf("-1
");
19 
20 
21     }
22     return 0;
23 }
View Code

HDU 6301

Problem Description

Chiaki has an array of jholds.
Chiaki would like to find a lexicographically minimal array which meets the facts.
 

Input

There are multiple test cases. The first line of input contains an integer 6.
 

Output

For each test case, output n integers denoting the lexicographically minimal array. Integers should be separated by a single space, and no extra spaces are allowed at the end of lines.
 

Sample Input

3
2 1
1 2
4 2
1 2
3 4
5 2
1 3
2 4
 

Sample Output

1 2
1 2 1 2
1 2 3 1 1
 

 题解：

先对给定的区间排序，然后每次讲前一个比后一个区间多出的部分加入优先对列，找到最小的然后放到ans数组。这个代码有点卡时间，

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 
 5 const int MAXN = 1e5+10;
 6 
 7 
 8 int t,n,m;
 9 
10 int ans[MAXN];
11 struct node{
12     int l,r;
13 }p[MAXN],pre;
14 struct CMP{
15     bool operator ()(int a,int b){
16         return a > b;
17     }
18 };
19 bool cmp(node a,node b){
20     if(a.l!=b.l)    return a.l < b.l;
21     else return a.r > b.r;
22 }
23 
24 int main(){
25     scanf("%d",&t);
26     while(t--){
27         memset(ans,0,sizeof(ans));
28         memset(p,0,sizeof(p));
29         scanf("%d%d",&n,&m);
30         priority_queue<int, vector<int>,CMP>Q;
31         for(int i=0;i<m;i++){
32             scanf("%d%d",&p[i].l,&p[i].r);
33         }
34         sort(p,p+m,cmp);
35         int maxx = 1;
36         pre.l = p[0].l;
37         pre.r = p[0].r;
38         for(int i=1;i<=p[0].r-p[0].l+1;i++){
39             ans[p[0].l+i-1] = i;
40             maxx = max(maxx,i);
41         }
42         for(int i=1;i<m;i++){
43             if(p[i].r <= pre.r) continue;
44             else {
45                 for(int j=pre.l;j<min(p[i].l,pre.r+1);j++){
46                     Q.push(ans[j]);
47                 }
48                 int temp =max(pre.r+1,p[i].l);
49                 while(!Q.empty()&&temp<=p[i].r){
50                     ans[temp++]=Q.top();
51                     Q.pop();
52                 }
53                 if(temp<=p[i].r){
54                     for(int k=temp;k<=p[i].r;k++){
55                         ans[k]=++maxx;
56                     }
57                 }
58             }
59             pre.l=p[i].l;
60             pre.r=p[i].r;
61         }
62         for(int i=1;i<=n;i++){
63             if(ans[i]==0)ans[i]=1;
64         }
65         for(int i=1;i<n;i++){
66             printf("%d ",ans[i]);
67         }printf("%d
",ans[n]);
68     }
69     return 0;
70 }
View Code

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