CodeForces - 978 C. Letters

C. Letters
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are nn dormitories in Berland State University, they are numbered with integers from 11 to nn. Each dormitory consists of rooms, there are aiai rooms in ii-th dormitory. The rooms in ii-th dormitory are numbered from 11 to aiai.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all nn dormitories is written on an envelope. In this case, assume that all the rooms are numbered from 11 to a1+a2+⋯+ana1+a2+⋯+an and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case n=2n=2, a1=3a1=3 and a2=5a2=5 an envelope can have any integer from 11 to 88 written on it. If the number 77 is written on an envelope, it means that the letter should be delivered to the room number 44 of the second dormitory.
For each of mm letters by the room number among all nn dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
Input
The first line contains two integers nn and mm (1≤n,m≤2⋅105)(1≤n,m≤2⋅105) — the number of dormitories and the number of letters.
The second line contains a sequence a1,a2,…,ana1,a2,…,an (1≤ai≤1010)(1≤ai≤1010), where aiai equals to the number of rooms in the ii-th dormitory. The third line contains a sequence b1,b2,…,bmb1,b2,…,bm (1≤bj≤a1+a2+⋯+an)(1≤bj≤a1+a2+⋯+an), where bjbj equals to the room number (among all rooms of all dormitories) for the jj-th letter. All bjbj are given in increasing order.
Output
Print mm lines. For each letter print two integers ff and kk — the dormitory number ff (1≤f≤n)(1≤f≤n) and the room number kk in this dormitory (1≤k≤af)(1≤k≤af) to deliver the letter.
Examples
input
Copy
3 6
10 15 12
1 9 12 23 26 37

output
Copy
1 1
1 9
2 2
2 13
3 1
3 12

input
Copy
2 3
5 10000000000
5 6 9999999999

output
Copy
1 5
2 1
2 9999999994

Note
In the first example letters should be delivered in the following order:
  • the first letter in room 11 of the first dormitory
  • the second letter in room 99 of the first dormitory
  • the third letter in room 22 of the second dormitory
  • the fourth letter in room 1313 of the second dormitory
  • the fifth letter in room 11 of the third dormitory
  • the sixth letter in room 1212 of the third dormitory

  • 알고리즘 분석:
    제목:
    n개 기숙사, 각 기숙사는 a[i]개의 방, m통의 편지, b[i]가 이 방 번호에 대응합니다.
    분석:
    간단한 시뮬레이션만 하면 돼요. 처음에 이중 순환이 시간을 초과했지만 저는 많이 최적화되었어요. 그런데 여전히 시간을 초과했어요. 그냥 조금 바꿨는데 시간이 지났어요. 좀 화가 났어요.
    코드 구현:
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    #include  
    using namespace std;  
    const double eps = 1e-8;  
    typedef long long LL;  
    typedef unsigned long long ULL;  
    const int INF = 0x3f3f3f3f;  
    const int INT_M_INF = 0x7f7f7f7f;  
    const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
    const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
    const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
    const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};  
    const int MOD = 1e9 + 7;  
    const double pi = acos(-1.0);  
    const int MAXN=5010;  
    const int MAXM=100010;
    using namespace std;
    long long  a[200005],b[200005];
    int main()
    {
       
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
    	{     a[0]=0;              
    		for(int i=1;i<=n;i++)
    		{
    			scanf("%lld",&a[i]);
    			a[i]+=a[i-1];
    		}	
    		for(int i=1;i<=m;i++)
    			scanf("%lld",&b[i]);
    			int i=1;
    	    for(int j=1;j<=n;)
    		  {
    		  	
    				if(a[j]>=b[i])
    				{
    					printf("%d %lld
    ",j,b[i]-a[j-1]); i++; } else { j++; } if(i>m) break; } } return 0; }

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