Fzu 2171 defensive position II

E-defensive position II
Time Limit: 3000 Ms memory limit: 32768kb 64bit Io format: % i64d & % i64u
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Status
Description
There are a total of N soldiers in the army, each of which has its own capacity index XI,
During a drill, the headquarters determined m locations to be defended,
The headquarters will select M soldiers to go to the designated location for defensive tasks in turn,
The obtained reference index is the sum of the capabilities of M soldiers.

Over time, the headquarters will issue Q commands to replace m of soldiers defending,
The capability index of each soldier who completes the defensive task will drop by 1 due to fatigue and other reasons.

Now the soldiers are in a row. Please calculate the reference index of each defender.

Input
The input contains multiple groups of data.

The first line contains two integers, n, m, and Q (1 <=n <= 100000,1 <= m <= 100000 ,1 <= q <= ),
The N integers in the second row indicate the capability index XI (1 <= xi <= 1000) of each soldier ).

Next, row Q contains an integer x,
In the original queue, M soldiers starting with X are replaced with previous soldiers for defense. (1 <= x <= N-M + 1)

Output
Output Q rows, each row has an integer, which is the reference index of soldiers who defend each command execution.

Sample Input
5 3 3
2 1 3 1 4
1
2
3
Sample output
6
3
5

 

 

At the beginning, I saw a little bit of excitement. The result array was not opened four times. =

 

I have asked Wang daemon before. It is time-out to use the line segment tree to modify the interval.

At present, it seems that only templates can be set and cannot be used smoothly.

 

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define ll __int64#define MAXN 1000#define INF 0x7ffffff#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int sum[400000+10];// n<<2!!!!!int add[1000000];void pushup(int rt){    //if(sum[rt<<1]<0) sum[rt<<1]=0;    //if(sum[rt<<1|1]<0) sum[rt<<1|1]=0;    sum[rt]=sum[rt<<1] +sum[rt<<1|1];}void pushdown(int rt,int m){    if(add[rt])    {        add[rt<<1]+=add[rt];        add[rt<<1|1]+=add[rt];        sum[rt<<1]+=add[rt]*(m-(m>>1));        sum[rt<<1|1]+=add[rt]*(m>>1);        //if(sum[rt<<1]<0) sum[rt<<1]=0;        //if(sum[rt<<1|1]<0) sum[rt<<1|1]=0;        add[rt]=0;    }}void build(int l,int r,int rt){    if(l==r)    {        scanf("%d",&sum[rt]);        return ;    }    int m=(l+r)>>1;    build(lson); build(rson); pushup(rt);}void update(int L,int R,int c,int l,int r,int rt){    if(L<=l&&r<=R)    {        add[rt]+=c;        sum[rt]+=c*(r-l+1);        return ;    }    pushdown(rt,r-l+1);    int m=(l+r)>>1;    if(R<=m) update(L,R,c,lson);    else if(L>m) update(L,R,c,rson);    else    {        update(L,R,c,lson);        update(L,R,c,rson);    }    pushup(rt);}int query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return sum[rt];    }    pushdown(rt,r-l+1);    int m=(l+r)>>1;    int ret=0;    if(R<=m) ret+=query(L,R,lson);    else if(L>m) ret+=query(L,R,rson);    else    {        ret+=query(L,R,lson);        ret+=query(L,R,rson);    }    return ret;}int main(){    int i,j,n,m,q,wen,jian;    while(scanf("%d%d%d",&n,&m,&q)!=EOF)    {        mem(add,0);        build(1,n,1);        for(i=1;i<=q;i++)        {                scanf("%d",&wen);            printf("%d\n",query(wen,wen+m-1,1,n,1));            update(wen,wen+m-1,-1,1,n,1);        }    }    return 0;}