Codeforces Round #486 (Div. 3) C "Equal Sums" （map+pair<>）

Portal

 

• the meaning of problems

　　You k sequences, the i-th sequence containing n- _i elements;

　　Ask if there are two different sequences i, j;

　　The sum of i satisfying remove a sequence number is equal to the sum of the sequence j deleted after a certain number;

• problem solution

　　定义 map<int , pair<int ,int > >f;

　　For convenience of description, the above definition to map <x, pair <y, z>> f;

　　It represents the sum of the sequence y deleted after the number of first z x;

　　For the i-th sequence, after deleting the first number j, determines the sum cur after deletion;

　　F cur determining whether there is, if there is output directly, whereas keep looking;

•Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define pii pair<int ,int >
 4 const int maxn=2e5+50;
 5 
 6 int k;
 7 int a[maxn];
 8 map<int ,pii >f;
 9 
10 int main()
11 {
12     scanf("%d",&k);
13     for(int i=1;i <= k;++i)
14     {
15         int n;
16         scanf("%d",&n);
17 
18         int sum=0;
19         for(int j=1;j <= n;++j)
20         {
21             scanf("%d",a+j);
22             sum += a[j];
23         }
24 
25         for(int j=1;j <= n;++j)
26         {
27              int CUR sum- = A [J];
 28  
29              /// sequence before determining whether there is a CUR
 30              /// ! F [CUR] .first is necessary I =
 31              /// test sample can be understood as follows 
32              / * *
 33 is                  . 1
 34 is                  2
 35                  . 1. 1
 36              * / 
37 [              IF (! f.count (CUR) && F [CUR] .first = I)
 38 is              {
 39                  the puts ( " YES " );
 40                  PII tmp = F [ CUR];
 41 is                  the printf ("%d %d\n",i,j);
42                 printf("%d %d\n",tmp.first,tmp.second);
43                 return 0;
44             }
45             f[cur]=pii(i,j);
46         }
47     }
48     puts("NO");
49 
50     return 0;
51 }

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