BestCoder Round #32

时间：2015-03-15 00:43:15 阅读：129 评论：0 收藏：0 [点我收藏+]

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PM2.5

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 613 Accepted Submission(s): 326

Problem Description

Nowadays we use content of PM2.5 to discribe the quality of air. The lower content of PM2.5 one city have, the better quality of air it have. So we sort the cities according to the content of PM2.5 in asending order.

Sometimes one city’s rank may raise, however the content of PM2.5 of this city may raise too. It means that the quality of this city is not promoted. So this way of sort is not rational. In order to make it reasonable, we come up with a new way to sort the cityes. We order this cities through the diffrence between twice measurement of content of PM2.5 (first measurement – second measurement) in descending order, if there are ties, order them by the second measurement in asending order , if also tie, order them according to the input order.

Input

Multi test cases (about

Output

For each case, output the cities’ id in one line according to their order.

Sample Input

2 100 1 1 2 3 100 50 3 4 1 2

Sample Output

0 1 0 2 1

Source

BestCoder Round #32

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cstdlib>
 6 using namespace std;
 7 struct node
 8 {
 9     int num,f,s,c;
10 };
11 node city[107];
12 int ans[107];
13 int n;
14 bool cmp(node n1,node n2)
15 {
16     if(n1.c!=n2.c) return n1.c>n2.c;
17      if(n1.s!=n2.s) return n1.s<n2.s;
18     return n1.num<n2.num;
19 }
20 int main()
21 {
22   //  freopen("in.txt","r",stdin);
23     while(~scanf("%d",&n)){
24         memset(city,0,sizeof(city));
25         for(int i=0;i<n;i++)
26         {
27             city[i].num=i;
28             scanf("%d%d",&city[i].f,&city[i].s);
29             city[i].c=city[i].f-city[i].s;
30         }
31         sort(city,city+n,cmp);
32         for(int i=0;i<n;i++)
33         {
34             ans[city[i].num]=i;
35         }
36         bool ff=1;
37         for(int i=0;i<n;i++)
38         {
39             if(ff) printf("%d",city[i].num),ff=0;
40             else {
41                 printf(" %d",city[i].num);
42             }
43         }
44         printf("\n");
45     }
46     return 0;
47 }

Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2453 Accepted Submission(s): 605

Problem Description

When given an array

Input

Multi test cases. In the first line of the input file there is an integer

Output

For each case，the output should occupies exactly one line. The output format is Case #id: ans, here id is the data number starting from 1; ans is “Yes.” or “No.” (without quote) according to whether you can find

Sample Input

2 1 1 1 2 1 -1 0

Sample Output

Case #1: Yes. Case #2: No.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<set>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<vector>
 8 using namespace std;
 9 const int mod=1000007;
10 const int maxn=1000010;
11 const int add=1000000000;
12 long long sum1[maxn],sum2[maxn];
13 int n,k;
14 struct HASHMAP
15 {
16     int head[maxn],next[maxn],sum;
17     long long value[maxn];
18     void init()
19     {
20         memset(head,-1,sizeof(head));
21         sum=0;
22     }
23     bool check(long long val)
24     {
25         int h=(val%mod+mod)%mod;
26         for(int i=head[h];~i;i=next[i]){
27             if(value[i]==val) return 1;
28         }
29         return 0;
30     }
31     bool insert(long long val)
32     {
33         int h=(val%mod+mod)%mod;
34         for(int i=head[h];~i;i=next[i])
35             if(value[i]==val) return 1;
36         value[sum]=val;
37         next[sum]=head[h];
38         head[h]=sum++;
39         return 0;
40     }
41 }H1,H2;
42 int main()
43 {
44  //  freopen("in.txt","r",stdin);
45     int t;
46     scanf("%d",&t);
47      int value;
48      int cnt=1;
49     while(t--){
50         H1.init();H2.init();
51         memset(sum1,0,sizeof(sum1));
52          int sign1=1,sign2=1;
53          scanf("%d%d",&n,&k);
54          bool ans=0;
55          int tt;
56         H1.insert(0);
57           for(int i=1;i<=n;i++)
58           {
59               scanf("%d",&value);
60               if(ans==1) continue;
61              if(sign1==1) sum1[i]=sum1[i-1]+value;
62              else sum1[i]=sum1[i-1]-value;
63              if(i%2==0) H1.insert(sum1[i]);
64               if(H1.check(sum1[i]-k)) ans=1;
65               sum2[i]=-sum1[i];
66                if(i%2) H2.insert(sum2[i]);
67               if(H2.check(sum2[i]-k)) ans=1;
68               sign1*=-1;
69           }
70         if(ans) printf("Case #%d: Yes.\n",cnt++);
71         else printf("Case #%d: No.\n",cnt++);
72     }
73     return 0;
74 }

BestCoder Round #32

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原文地址：http://www.cnblogs.com/codeyuan/p/4338597.html

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