标签：problem ted nis finish scanf codeforce result space nbsp
While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array a and integer x. He should find the number of different ordered pairs of indexes (i,?j) such that ai?≤?aj and there are exactly k integers y such that ai?≤?y?≤?aj and y is divisible by x.
In this problem it is meant that pair (i,?j) is equal to (j,?i) only if i is equal to j. For example pair (1,?2) is not the same as (2,?1).
Input
The first line contains 3 integers n,?x,?k (1?≤?n?≤?105,?1?≤?x?≤?109,?0?≤?k?≤?109), where n is the size of the array a and x and k are numbers from the statement.
The second line contains n integers ai (1?≤?ai?≤?109) — the elements of the array a.
Output
Print one integer — the answer to the problem.
Example
4 2 1
1 3 5 7
3
4 2 0
5 3 1 7
4
5 3 1
3 3 3 3 3
25
Note
In first sample there are only three suitable pairs of indexes — (1,?2),?(2,?3),?(3,?4).
In second sample there are four suitable pairs of indexes(1,?1),?(2,?2),?(3,?3),?(4,?4).
In third sample every pair (i,?j) is suitable, so the answer is 5?*?5?=?25.
#include <cstdio> #include <iostream> #include <string.h> #include <cmath> #include <algorithm> #define LL long long using namespace std; LL a[100000+10]; int main() { int n,x,k; LL result = 0; while(scanf("%d",&n)!=EOF) { scanf("%d%d",&x,&k); result = 0; for(int i=1;i<=n;i++)cin>>a[i]; sort(a+1,a+n+1); for(int i=1;i<=n;i++) { LL l = max(a[i],(k+((a[i]-1)/x))*x); LL r = max(a[i],(k+1+((a[i]-1)/x))*x); //cout<<l; //printf(" "); //cout<<r<<endl; result += (lower_bound(a+1,a+n+1,r)-lower_bound(a+1,a+n+1,l)); } printf("%lld\n",result); } return 0; } //2 6 13 6-(3-1) //6 8 10 12 //(k+((a[i]-1)/x))*x=a[j]; //a[j]/x-(a[i]-1)/x == k+1?
标签：problem ted nis finish scanf codeforce result space nbsp
原文地址：http://www.cnblogs.com/--lr/p/7922647.html
