The little girl loves the problems on array queries very much.
One day she came across a rather well-known problem: you’ve got an array of n elements (the elements of the array are indexed starting from 1); also, there are q queries, each one is defined by a pair of integers li, ri (1≤li≤ri≤n). You need to find for each query the sum of elements of the array with indexes from li to ri, inclusive.
The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum. Input
The first line contains two space-separated integers n (1≤n≤2·105) and q (1≤q≤2·105) — the number of elements in the array and the number of queries, correspondingly.
The next line contains n space-separated integers ai (1≤ai≤2·105) — the array elements.
Each of the following q lines contains two space-separated integers li and ri (1≤li≤ri≤n) — the i-th query. Output
In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Sample test(s) Input
3 3 5 3 2 1 2 2 3 1 3
Output
25
Input
5 3 5 2 4 1 3 1 5 2 3 2 3
Output
33
贪心，，统计出每一个位置出现的次数，然后当然出现最多的那个位置放最大的数，然后相乘累加就行
/*************************************************************************> File Name: CF-169-C.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年04月02日 星期四 15时11分23秒 ************************************************************************/;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;LL;typedef pair <int, int> PLL;N = 200100;int sum[N];int arr[N];int main(){int n, q;while (~scanf(“%d%d”, &n, &q)){memset(sum, 0, sizeof(sum));for (int i = 1; i <= n; ++i){scanf(“%d”, &arr[i]);}int l, r;while (q–){scanf(“%d%d”, &l, &r);++sum[l];–sum[r + 1];}for (int i = 1; i <= n; ++i){sum[i] += sum[i – 1];}sort(sum + 1, sum + n + 1);sort(arr + 1, arr + n + 1);LL ans = 0;for (int i = n; i >= 1; –i){ans += (LL)arr[i] * sum[i];}printf(“%I64d\n”, ans);}return 0;}
怠惰是贫穷的制造厂。
