| Time Limit: 404MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
Here we go!
Let‘s define the diversity of a list of numbers to be the difference between the largest and smallest number in the list.
For example, the diversity of the list (1, -1, 2, 7) = 7 - (-1) = 8.
A substring of a list is considered a non-empty sequence of contiguous numbers from the list. For example, for the list (1,3,7), the substrings are (1), (3), (7), (1,3), (3,7), (1,3,7). A subsequence of a list is defined to be a non-empty sequence of numbers obtained by deleting some elements from the list. For example, for the list (1,3,7), the subsequences are (1), (3), (7), (1,3), (3,7), (1,7), (1,3,7).
Given a list of length N find the number of substrings and subsequences in this list with the maximum diversity. If a substring/subsequence having maximum diversity occurs multiple times in the list, each of its occurances adds towards the answer. And tell Harry Potter your answer.
Input (STDIN):
The first line contains T, the number of test cases. Then follow T test case blocks.
Each blocks starts with the first line containing the number N.
The second line contains a list of numbers in this list.
Output (STDOUT):
For each test case, output the number of substrings and the number of subsequences in this list with the maximum diversity.
Since the answers maybe very large, output them modulo 1000000007.
Constraints:
T <= 10
N <= 100,000
Each number in the list is between 1 and 100,000 inclusive.
Time Limit: 2 s
Memory Limit: 32 MB
Sample Input:
3
3
1 2 3
4
1 4 3 4
3
3 2 1
Sample Output:
1 2
3 6
12
Enough with this Harry Potter, please! What are we, twelve-year olds? Let‘s get our teeth into some real pumpkin pasties -- oops, programming problems!
Here we go!
Let‘s define the diversity of a list of numbers to be the difference between the largest and smallest number in the list.
For example, the diversity of the list (1, -1, 2, 7) = 7 - (-1) = 8.
A substring of a list is considered a non-empty sequence of contiguous numbers from the list. For example, for the list (1,3,7), the substrings are (1), (3), (7), (1,3), (3,7), (1,3,7). A subsequence of a list is defined to be a non-empty sequence of numbers obtained by deleting some elements from the list. For example, for the list (1,3,7), the subsequences are (1), (3), (7), (1,3), (3,7), (1,7), (1,3,7).
Given a list of length N find the number of substrings and subsequences in this list with the maximum diversity. If a substring/subsequence having maximum diversity occurs multiple times in the list, each of its occurences adds towards the answer. And tell Harry Potter your answer
Input (STDIN):
The first line contains T, the number of test cases. Then follow T test case blocks.
Each blocks starts with the first line containing the number N.
The second line contains a list of numbers in this list.
Output (STDOUT):
For each test case, output the number of substrings and the number of subsequences in this list with the maximum diversity.
Since the answers maybe very large, output them modulo 1000000007.
Constraints:
T <= 10
N <= 100,000
Each number in the list is between 1 and 100,000 inclusive.
Sample Input:
3
3
1 2 3
4
1 4 3 4
3
3 2 1
Sample Output:
1 2
3 6
1 2
/**
题意 ：给你一个串，问使得串中的最大值 - 最小值 为 deliver
然后看有多少个substring 和 subsequence 串 是的 deliver 与
原串相等
做法 ：
对于一个串，找到最大值mmax，最小值进行标记mmin，然后看分别由多少个
mmax 由_mmax记录 ,mmin 由_mmin 记录
然后符合要求的subsequence是
（容斥原理 )包含最大值和最小值的子集的个数 = 总的子集个数 - 只有最小值的子集个数 - 只有最大值的子集的个数 + 既没有最小值又没有最大值的子集的个数
符合要求的substring 是
从0开始枚举
有t1 标记离当前位置最近的mmin下标 ,用t2标记离当前位置最近的mmax下标
然后进行枚举
sum = sum + mmin(t1 +1,t2+1);
PS:当 mmin == mmax 时要特别处理
substring 是 (n*(n+1))/2;
subsequence 是 2^n - 1
**/
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <stdio.h>
using namespace std;
#define maxn 100000 + 100
#define mod 1000000007
long long mmap[maxn];
long long _next[maxn];
int main()
{
_next[0] = 1;
for(int i = 1; i < maxn; i++)
{
_next[i] = _next[i - 1] * 2 % mod;
}
int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);
long long mmin = 0xfffffff;
long long mmax = 0;
for(int i = 0; i < n; i++)
{
scanf("%lld", &mmap[i]);
mmin = min(mmin, mmap[i]);
mmax = max(mmax, mmap[i]);
}
long long sum1 = 0;
long long sum2 = 0;
if(mmax == mmin)
{
sum1 = ((n * (n + 1)) / 2) % mod;
sum2 = _next[n] - 1;
printf("%lld %lld\n", sum1, sum2);
continue;
}
int _mmin = 0;
int _mmax = 0;
int t1 = -1, t2 = -1;
for(int i = 0; i < n; i++)
{
if(mmap[i] == mmin) {
t1 = i;
_mmin ++;
}
if(mmap[i] == mmax) {
t2 = i;
_mmax ++;
}
sum1 = (sum1 + min(t1 + 1, t2 + 1)) % mod;
}
sum2 = (_next[n] - _next[n - _mmin] - _next[n - _mmax] + _next[n - _mmax - _mmin]) % mod;
if(sum2 < 0) {
sum2 += mod;
}
printf("%lld %lld\n", sum1, sum2);
}
return 0;
}