Ultra-QuickSort
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
Input
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input
5 9 1 0 5 4 3 1 2 3 0
Sample Output
6 0 题目是求每个数前面比它大的数量的总和，就是逆对数了。只要是离散化就可以做了。i-query(index[i])：当前是对i个数，query(index[i])是它前面比它小的数，减去它就是前面比它大的数量了。
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <algorithm>
5 #define lowbit(x) x&(-x)
6 #define ll long long
7 using namespace std;
8 const int MAX = 5e5+5;
9 int n, tree[MAX],index[MAX];
10 struct Nod{
11 int num,id;
12 }nod[MAX];
13 bool cmp(Nod a, Nod b){
14 return a.num < b.num;
15 }
16 void add(int x, int y){
17 while(x < MAX){
18 tree[x] += y;
19 x += lowbit(x);
20 }
21 }
22
23 int query(int x){
24 int sum = 0;
25 while(x > 0){
26 sum += tree[x];
27 x -= lowbit(x);
28 }
29 return sum;
30 }
31 int main(){
32 while(scanf("%d",&n)&&n){
33 for(int i = 1; i <= n; i ++){
34 scanf("%d",&nod[i].num);
35 nod[i].id = i;
36 }
37 sort(nod+1,nod+1+n,cmp);
38 for(int i = 1; i <= n; i ++)index[nod[i].id] = i;
39 memset(tree,0,sizeof(tree));
40 ll ans = 0;
41 for(int i = 1; i <= n; i ++){
42 add(index[i],1);
43 //printf("%d+++++%d\n",index[i],query(index[i]));
44 ans += i-query(index[i]);
45 }
46 printf("%lld\n",ans);
47 }
48 }
Enemy is weak
The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep".
Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.
In Shapur‘s opinion the weakness of an army is equal to the number of tripletsi,?j,?k such that i?<?j?<?k and ai?>?aj?>?ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.
Help Shapur find out how weak the Romans are.
Input
The first line of input contains a single number n (3?≤?n?≤?106) — the number of men in Roman army. Next line contains n different positive integers ai (1?≤?i?≤?n,?1?≤?ai?≤?109) — powers of men in the Roman army.
Output
A single integer number, the weakness of the Roman army.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Example
Input
33 2 1
Output
1
Input
32 3 1
Output
0
Input
410 8 3 1
Output
4
Input
41 5 4 3
Output
1 这个就是求三个递增的数量，也可以看成第i个数，求前面比它大的数量和后面比它小的数量。下面k是前面比它大的数量，求后面比它小的数量：由于k是第i个数前面比第i个数大的数量，所以后面比第i个数小的数量 为(n-index[i])-(i-1-k)就是n-i-index[i]+k+1,(n-index[i])是比第i个数小的 总数，把它减去前面比第i个数小的数量就是后面的数量了，前面比第i个数小的 数量为(i-1-k),所以(n-index[i])-(i-1-k)就推来了
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <algorithm>
5 #define lowbit(x) x&(-x)
6 #define ll long long
7 using namespace std;
8 const int MAX = 1e6+10;
9 int n,tree[MAX],index[MAX];
10 int nex[MAX],pre[MAX];
11 struct Nod{
12 int val,id;
13 }nod[MAX];
14 bool cmp(Nod a, Nod b){
15 return a.val > b.val;
16 }
17 void add(int x){
18 while(x < MAX){
19 tree[x] ++;
20 x += lowbit(x);
21 }
22 }
23 int query(int x){
24 int sum = 0;
25 while(x > 0){
26 sum += tree[x];
27 x -= lowbit(x);
28 }
29 return sum;
30 }
31 int main(){
32 scanf("%d",&n);
33 for(int i = 1; i <= n; i ++)scanf("%d",&nod[i].val),nod[i].id=i;
34 sort(nod+1,nod+1+n,cmp);
35 for(int i = 1; i <= n; i ++)index[nod[i].id] = i;
36 ll ans = 0,k;
37
38 for(int i = 1; i <= n; i ++){
39 k = query(index[i]);
40 ans += k*(n-i-index[i]+k+1);
41 add(index[i]);
42 }
43 cout << ans << endl;
44 return 0;
45 }