D. Frets On Fire
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Miyako came to the flea kingdom with a ukulele. She became good friends with local flea residents and played beautiful music for them every day.
In return, the fleas made a bigger ukulele for her: it has nn strings, and each string has (1018+1)(1018+1) frets numerated from 00 to 10181018. The fleas use the array s1,s2,…,sns1,s2,…,sn to describe the ukulele's tuning, that is, the pitch of the jj-th fret on the ii-th string is the integer si+jsi+j.
Miyako is about to leave the kingdom, but the fleas hope that Miyako will answer some last questions for them.
Each question is in the form of: "How many different pitches are there, if we consider frets between ll and rr (inclusive) on all strings?"
Miyako is about to visit the cricket kingdom and has no time to answer all the questions. Please help her with this task!
Formally, you are given a matrix with nn rows and (1018+1)(1018+1) columns, where the cell in the ii-th row and jj-th column (0≤j≤10180≤j≤1018) contains the integer si+jsi+j. You are to answer qq queries, in the kk-th query you have to answer the number of distinct integers in the matrix from the lklk-th to the rkrk-th columns, inclusive.
Input
The first line contains an integer nn (1≤n≤1000001≤n≤100000) — the number of strings.
The second line contains nn integers s1,s2,…,sns1,s2,…,sn (0≤si≤10180≤si≤1018) — the tuning of the ukulele.
The third line contains an integer qq (1≤q≤1000001≤q≤100000) — the number of questions.
The kk-th among the following qq lines contains two integers lklk，rkrk (0≤lk≤rk≤10180≤lk≤rk≤1018) — a question from the fleas.
Output
Output one number for each question, separated by spaces — the number of different pitches.
Examples
input
Copy
6
3 1 4 1 5 9
3
7 7
0 2
8 17
output
Copy
5 10 18
input
Copy
2
1 500000000000000000
2
1000000000000000000 1000000000000000000
0 1000000000000000000
output
Copy
2 1500000000000000000
题意
有n个数，m次询问。看样例，a_i为第一列，后面递增，组成矩阵，问从第l列到第r列有多少个不同的数
思路
每一行询问的区间为[a_i + l, a_i + r]，有些区间会重叠在一起，有些区间不会，所以先计算出所有相邻的两个数的差，排序，按r-l+1二分查，把这些差分为两部分：后面为不重叠的，前面为重叠的。不重叠部分都*(r-l+1)，重叠部分算出差的前缀和+(r-l+1)。代码实现上t多减了一个1，即把重叠部分的（r-l+1)跟不重叠部分一起算了。
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 5;
const int MOD = 1e9 + 9;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define F(i, l, r) for(int i = l;i <= (r);++i)
#define RF(i, l, r) for(int i = l;i >= (r);--i)
LL m, n, a[N], b[N], c[N];
int main()
{
//freopen("in.txt", "r", stdin);
cin >> n;
F(i, 1, n) cin >> a[i];
sort(a + 1, a + n + 1);
F(i, 1, n - 1) b[i] = a[i + 1] - a[i];
sort(b + 1, b + n);
F(i, 1, n - 1) c[i] = c[i - 1] + b[i];//前缀和
cin >> m;
while(m--)
{
LL l, r;
cin >> l >> r;
LL t = lower_bound(b + 1, b + n, r - l + 1) - b - 1;
cout << (r - l + 1) * (n - t) + c[t] << " ";
}
return 0;
}