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 100), every case contains an integer n which represents there are n cities to be sorted in the first line.
Cities are numbered through 0 to n−1.
In the next n lines each line contains two integers which represent the first and second measurement of content of PM2.5
The ith line describes the information of city i−1
Please process to the end of file.
[Technical Specification]
all integers are in the range [1,100]
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
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 (a0,a1,a2,?an−1) and an integer K, you are expected to judge whether there is a pair (i,j)(0≤i≤j<n) which makes that NP−sum(i,j) equals to K true. Here NP−sum(i,j)=ai−ai+1+ai+2+?+(−1)j−iaj
Input
Multi test cases. In the first line of the input file there is an integer T indicates the number of test cases.
In the next 2∗T lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers n and K which are mentioned above.
The second line contain (a0,a1,a2,?an−1)separated by exact one space.
[Technical Specification]
All input items are integers.
0<T≤25,1≤n≤1000000,−1000000000≤ai≤1000000000,−1000000000≤K≤1000000000
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 (i,j) which makes PN−sum(i,j) equals to K.
See the sample for more details.
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 }