pre[i]第i位数往前走多少位碰到和它相同的数
dp[i]表示长度为i的子串,dp[i]可以由dp[i-1]加上从i到n的pre[i]>i-1的数减去最后一段长度为i-1的断中的不同的数得到….
爆int+有点卡内存….
SubstringsTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2300Accepted Submission(s): 716
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
71 1 2 3 4 4 531230
Sample Output
71012
Source
/* ***********************************************Author:CKbossCreated Time :2015年08月17日 星期一 22时06分06秒File Name:HDOJ4455.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef long long int LL;const int maxn=1001000;int n;LL dp[maxn];int a[maxn];int pre[maxn];int spre[maxn];int wz[maxn];bool vis[maxn];/*************BIT*********************/inline int lowbit(int x) { return x&(-x); }int tree[maxn];void add(int p,int v){for(int i=p;i<maxn;i+=lowbit(i))tree[i]+=v;}LL sum(int p){LL ret=0;for(int i=p;i;i-=lowbit(i)) ret+=tree[i];return ret;}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%d",&n)!=EOF&&n){for(int i=1;i<=n;i++) scanf("%d",a+i);memset(wz,-1,sizeof(wz));memset(pre,0,sizeof(pre));memset(spre,0,sizeof(spre));memset(tree,0,sizeof(tree));memset(vis,false,sizeof(vis));for(int i=n;i>=1;i–){int x=a[i];if(wz[x]==-1) wz[x]=i;else{spre[wz[x]-i]++;pre[wz[x]]=wz[x]-i;wz[x]=i;}}int zero=0,nozero;for(int i=1;i<=n;i++){if(pre[i]==0) zero++;if(spre[i]) add(i,spre[i]);}int ed=n;int siz=0;nozero=n-zero;dp[1]=n; zero–;for(int i=2;i<=n;i++){if(vis[a[ed]]==false){vis[a[ed]]=true;siz++;}ed–;int B=siz;int A=zero+nozero-sum(i-1);dp[i]=dp[i-1]+A-B;if(pre[i]) nozero–,add(pre[i],-1);else zero–;}int Q;scanf("%d",&Q);while(Q–){int x;scanf("%d",&x);printf("%lld\n",dp[x]);}}return 0;}
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