杭电acm 4455(dp)
2017-08-10 22:01
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Substrings
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3320 Accepted Submission(s): 1024
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
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
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
Source
2012 Asia Hangzhou Regional Contest
题目大意:给一个长度为n的整数序列,定义egg(i,j)表示区间[i,j]中不同的数的个数。m次询问,每次询问x,表示求所有长度为x连续区间的 egg 之和。
题目分析:定义dp(len)表示所有长度为len的连续区间的egg之和。
则,dp(len)=dp(len-1)-egg(最后一个长度为len-1的连续区间)+每个长度为len的区间中的新增元素不重复的区间个数。
想法:
egg(最后一个长度为len-1的连续区间)用num【i】表示。
每个长度为len的区间中的新增元素不重复的区间个数用sum【i】表示。
用c【i】来统计有多少个元素与前面距离i的元素相等(或与距离i的元素为第一元素)。
用pre【i】来之前值为i的元素的位置。
把以上数组初始化后+ dp[i]=dp[i-1]-num[i-1]+sum[i];(动态转移方程)差不多了。
代码:
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
using namespace std;
const int maxn=1000050;
typedef long long LL;
int c[maxn],sum[maxn],num[maxn],pre[maxn];
LL dp[maxn];
int n,m,a[maxn];
int main()
{
while(scanf("%d",&n)&&n)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
memset(pre,0,sizeof(pre));
memset(c,0,sizeof(c));
for(int i=1;i<=n;i++)
{
c[i-pre[a[i]]]++;
pre[a[i]]=i;
}
sum
=c
;
for(int i=n-1;i>0;i--)
{
sum[i]=sum[i+1]+c[i];
}
memset(c,0,sizeof(c));//数组重用
c[a
]=1;
num[1]=1;
for(int i=2;i<=n;i++)
{
if(c[a[n-i+1]]==0)
{
num[i]=num[i-1]+1;
c[a[n-i+1]]=1;
}
else
{
num[i]=num[i-1];
}
}
dp[1]=n;
for(int i=2;i<=n;i++)
{
dp[i]=dp[i-1]-num[i-1]+sum[i];
}
scanf("%d",&m);
for(int i=0;i<m;i++)
{
int x;
scanf("%d",&x);
printf("%I64d\n",dp[x]);
}
}
return 0;
}
代码二:简化sum【】
#include<stdio.h>
#include<string.h>
# define LL long long
const int N=1000000;
int a[N+5];
int A[N+5];
int f[N+5];
int cnt[N+5];
LL dp[N+5];
int main()
{
int n;
while(scanf("%d",&n)&&n)
{
memset(f,0,sizeof(f));
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=n;i++)
{
scanf("%d",a+i);
cnt[i-f[a[i]]]++;
printf("%d\t",i-f[a[i]]);
f[a[i]]=i;
}
printf("\n");
for(int i=1;i<=n;i++)
{
printf("%d\t",cnt[i]);
}
printf("\n");
for(int i=1;i<=n;i++)
{
printf("%d\t",f[i]);
}
printf("\n");
memset(f,0,sizeof(f));
A[n+1]=0;
for(int i=n;i>=1;i--)
{
if(f[a[i]]) A[i]=A[i+1];
else A[i]=A[i+1]+1;
f[a[i]]=1;
}
dp[1]=n;
int sum=n;
for(int i=2;i<=n;i++)
{
dp[i]=dp[i-1]-A[n-i+2];
sum-=cnt[i-1];
dp[i]+=sum;
}
int m,x;
scanf("%d",&m);
while(m--)
{
scanf("%d",&x);
printf("%lld\n",dp[x]);
}
}
return 0;
}
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