Coder Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2226 Accepted Submission(s): 907 Problem Description In mathematics and computer science, an algorithm describes a set of procedures or instructions that define a procedure. The term has become increasing popular since the advent of cheap and reliable computers. Many companies now employ a single coder to write an algorithm that will replace many other employees. An added benefit to the employer is that the coder will also become redundant once their work is done. 1 You are now the signle coder, and have been assigned a new task writing code, since your boss would like to replace many other employees (and you when you become redundant once your task is complete). Your code should be able to complete a task to replace these employees who do nothing all day but eating: make the digest sum. By saying “digest sum” we study some properties of data. For the sake of simplicity, our data is a set of integers. Your code should give response to following operations: 1. add x – add the element x to the set; 2. del x – remove the element x from the set; 3. sum – find the digest sum of the set. The digest sum should be understood by where the set S is written as {a1, a2, ... , ak} satisfying a1 < a2 < a3 < ... < ak Can you complete this task (and be then fired)? ------------------------------------------------------------------------------ 1 See http://uncyclopedia.wikia.com/wiki/Algorithm Input There’re several test cases. In each test case, the first line contains one integer N ( 1 <= N <= 105 ), the number of operations to process. Then following is n lines, each one containing one of three operations: “add x” or “del x” or “sum”. You may assume that 1 <= x <= 109. Please see the sample for detailed format. For any “add x” it is guaranteed that x is not currently in the set just before this operation. For any “del x” it is guaranteed that x must currently be in the set just before this operation. Please process until EOF (End Of File). Output For each operation “sum” please print one line containing exactly one integer denoting the digest sum of the current set. Print 0 if the set is empty. Sample Input 9 add 1 add 2 add 3 add 4 add 5 sum add 6 del 3 sum 6 add 1 add 3 add 5 add 7 add 9 sum Sample Output 3 4 5 Hint C++ maybe run faster than G++ in this problem. Source 2012 ACM/ICPC Asia Regional Chengdu Online Recommend liuyiding 線段樹，對於這題，sum要用int64 ,這一點，我錯了好幾次， 這題要用線段樹，是很顯的啦，首先，我們用cnt保存這個區間上的結點的個數，用sum來保存，這個區間上的所有的mod i的和，這樣，我們可以得到維護
l[num].sum[i]=l[lson].sum[i]+l[rson].sum[((i-l[lson].cnt)%5+5)%5];
處理的時候，先要離線化，再用一個二分就好了！
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define lson (num<<1)
#define rson (num<<1|1)
#define N 100050
using namespace std;
struct node {
int cnt;
__int64 sum[5];
}l[N*20];
char str[N][10];int p[N],q[N],pq[N],no[N],ans;
bool cmp(int a,int b){return p[a]<p[b];}
bool cmp1(int a,int b){return a<b;}
int find(int goal)
{
int s,e,mid;
s=0,e=ans;
while(s<=e)
{
mid=(s+e)>>1;
if(no[mid]==goal)
return mid;
else if(no[mid]>goal)
e=mid;
else if(no[mid]<goal)
s=mid;
}
}
void build(int num,int s,int e)
{
l[num].cnt=0;
for(int i=0;i<5;i++)
l[num].sum[i]=0;
if(s>=e)
return;
int mid=(s+e)>>1;
build(lson,s,mid);
build(rson,mid+1,e);
}
int update(int num,int s,int e,int pos,int color,int k)
{
if(s>=e)
{
l[num].cnt=color;
l[num].sum[0]=color*k;
return 1;
}
int mid=(s+e)>>1;
if(pos<=mid)
update(lson,s,mid,pos,color,k);
else if(pos>mid)
update(rson,mid+1,e,pos,color,k);
l[num].cnt=l[lson].cnt+l[rson].cnt;
for(int i=0;i<5;i++)
{
l[num].sum[i]=l[lson].sum[i]+l[rson].sum[((i-l[lson].cnt)%5+5)%5];
}
}
int main ()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(ans=0,i=0;i<n;i++)
{
scanf("%s",&str[i]);
if(str[i][0]=='a')
{
scanf("%d",&p[i]);
no[ans]=p[i];
q[ans++]=i;
}
else if(str[i][0]=='d')
{
scanf("%d",&p[i]);
}
}
sort(q,q+ans,cmp);
sort(no,no+ans,cmp1);
build(1,1,ans);
for(i=0;i<ans;i++)
pq[q[i]]=i;
for(i=0;i<n;i++)
{
if(str[i][0]=='a')
{
update(1,1,ans,pq[i]+1,1,p[i]);
}
else if(str[i][0]=='d')
{
int pos=find(p[i]);
update(1,1,ans,pos+1,0,p[i]);
}
else if(str[i][0]=='s')
{
printf("%I64d\n",l[1].sum[2]);
}
}
}
return 0;
}
