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There are dormitories in Berland State University, they are numbered with integers from to . Each dormitory consists of rooms, there are rooms in -th dormitory. The rooms in -th dormitory are numbered from to .
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 dormitories is written on an envelope. In this case, assume that all the rooms are numbered from to 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 , and an envelope can have any integer from to written on it. If the number is written on an envelope, it means that the letter should be delivered to the room number of the second dormitory.
For each of letters by the room number among all dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
InputThe first line contains two integers and — the number of dormitories and the number of letters.
The second line contains a sequence , where equals to the number of rooms in the -th dormitory. The third line contains a sequence , where equals to the room number (among all rooms of all dormitories) for the -th letter. All are given in increasing order.
OutputPrint lines. For each letter print two integers and — the dormitory number and the room number in this dormitory to deliver the letter.
Examples3 6 10 15 12 1 9 12 23 26 37
1 1 1 9 2 2 2 13 3 1 3 12
2 3 5 10000000000 5 6 9999999999
1 5 2 1 2 9999999994
In the first example letters should be delivered in the following order:
题意:
直接分析数据吧。
给出n,m。有n个宿舍楼以及每个宿舍的房间数,把所有房间号按1到a1+a2+...+an编号。再给出m次询问,每次询问一个编号,问这个编号是第几个宿舍楼的第几个宿舍。m个询问按递增顺序给出。
解题思路:
因为按递增顺序给出,所以每找一次就可以存下当前找到的sum和对应的下标,下次找直接在这个基础上继续找下去就行了。
AC代码:
- #include<stdio.h>
-
- typedef long long ll;
-
- int main()
- {
- ll n,m;
- while(~scanf("%lld%lld",&n,&m))
- {
- ll a[n];
- for(ll i=0;i<n;i++)
- {
- scanf("%lld",&a[i]);
- }
- ll sum=0,x;
- int f=0,i;//记录下标避免重复查找
- while(m--)
- {
- scanf("%lld",&x);//按递增顺序给出
- x-=sum;
- for(i=f;i<n;i++)
- {
- if(x-a[i]>0)
- {
- x-=a[i];
- sum+=a[i];
- }
- else
- {
- printf("%d %lld\n",i+1,x);
- f=i;
- break;
- }
- }
- }
- }
- return 0;
- }
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