Substrings
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1727 Accepted Submission(s): 518
Problem Description
XXX has an array of length n. XXX wants to know that, for a given w, what is the sum of the distinct elements’ number in all substrings of length w. For example, the array is { 1 1 2 3 4 4 5 } When w = 3, there are five substrings of length 3. They are (1,1,2),(1,2,3),(2,3,4),(3,4,4),(4,4,5)
The distinct elements’ number of those five substrings are 2,3,3,2,2.
So the sum of the distinct elements’ number should be 2+3+3+2+2 = 12
The distinct elements’ number of those five substrings are 2,3,3,2,2.
So the sum of the distinct elements’ number should be 2+3+3+2+2 = 12
Input
There are several test cases.
Each test case starts with a positive integer n, the array length. The next line consists of n integers a1,a2…an, representing the elements of the array.
Then there is a line with an integer Q, the number of queries. At last Q lines follow, each contains one integer w, the substring length of query. The input data ends with n = 0 For all cases, 0<w<=n<=106, 0<=Q<=104, 0<= a1,a2…an <=106
Each test case starts with a positive integer n, the array length. The next line consists of n integers a1,a2…an, representing the elements of the array.
Then there is a line with an integer Q, the number of queries. At last Q lines follow, each contains one integer w, the substring length of query. The input data ends with n = 0 For all cases, 0<w<=n<=106, 0<=Q<=104, 0<= a1,a2…an <=106
Output
For each test case, your program should output exactly Q lines, the sum of the distinct number in all substrings of length w for each query.
Sample Input
7
1 1 2 3 4 4 5
3
1
2
3
0
Sample Output
7
10
12
1、
非常明显，长度为1的答案为dp[1]=n;
2、
长度为2的为dp[2]=dp[1]+x-y=7+4-1=10；
X为添加的一个数和前边不同的个数，{1,1}，{1,2}，{2,3}，{3,4}，{4,4}，{4,5} 为4；
Y为去掉的不足2的区间有几个不同数字，长度为1的最后一个区间{5},须要舍去。为1；
3、
长度为3的为dp[3]=dp[2]+x-y=10+4-2=12。
X为添加的一个数和前边不同的个数，{1,1,2}。{1,2,3}，{2,3,4}。{3,4,4}，{4,4,5}为4；
Y为去掉的不足3的区间有几个不同数字。长度为2的最后一个区间{4,5},须要舍去，为2；
所以，我们须要得到当前数字和它上次出现的位置差的大小。详细实现看代码。。
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1000005
#define LL __int64
const int inf=0x1f1f1f1f;
int a[N],len[N],pre[N],vis[N],f[N];
LL dp[N];
int main()
{
int i,n,m;
while(scanf("%d",&n),n)
{
memset(pre,-1,sizeof(pre)); //记录一个值上次出现的位置
memset(len,0,sizeof(len)); //len[i]：有几个间隔为i的数
memset(dp,0,sizeof(dp)); //记录终于答案
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
len[i-pre[a[i]]]++;
pre[a[i]]=i;
}
for(i=n-1;i>=0;i--)
len[i]+=len[i+1];
memset(f,0,sizeof(f)); //f[i]从后往前记录后i个数有几个不同值
memset(vis,0,sizeof(vis));
for(i=1;i<n;i++)
{
f[i]=f[i-1];
if(!vis[a[n-i]])
{
vis[a[n-i]]=1;
f[i]++;
}
}
dp[1]=n;
for(i=2;i<=n;i++)
dp[i]=dp[i-1]+len[i]-f[i-1];
scanf("%d",&m);
while(m--)
{
scanf("%d",&i);
printf("%I64d\n",dp[i]);
} }
return 0;
}