During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secret. After a pause, Dragon added:
— We have a contract. A rental agreement. He always works all day long. He likes silence. Besides that, there are many more advantages of living here in the Castle. Say, it is easy to justify a missed call: a phone ring can't reach the other side of the Castle from where the phone has been left. So, the imprisonment is just a tale. Actually, he thinks about everything. He is smart. For instance, he started replacing incandescent lamps with energy-saving lamps in the whole Castle...
Lpl chose a model of energy-saving lamps and started the replacement as described below. He numbered all rooms in the Castle and counted how many lamps in each room he needs to replace.
At the beginning of each month, Lpl buys mmm energy-saving lamps and replaces lamps in rooms according to his list. He starts from the first room in his list. If the lamps in this room are not replaced yet and Lpl has enough energy-saving lamps to replace all lamps, then he replaces all ones and takes the room out from the list. Otherwise, he'll just skip it and check the next room in his list. This process repeats until he has no energy-saving lamps or he has checked all rooms in his list. If he still has some energy-saving lamps after he has checked all rooms in his list, he'll save the rest of energy-saving lamps for the next month.
As soon as all the work is done, he ceases buying new lamps. They are very high quality and have a very long-life cycle.
Your task is for a given number of month and descriptions of rooms to compute in how many rooms the old lamps will be replaced with energy-saving ones and how many energy-saving lamps will remain by the end of each month.
Input
Each input will consist of a single test case.
The first line contains integers nnn and m(1≤n≤100000,1≤m≤100)m (1 \le n \le 100000, 1 \le m \le 100)m(1≤n≤100000,1≤m≤100) — the number of rooms in the Castle and the number of energy-saving lamps, which Lpl buys monthly.
The second line contains nnn integers k1,k2,...,knk_1, k_2, ..., k_nk1?,k2?,...,kn?
(1≤kj≤10000,j=1,2,...,n)(1 \le k_j \le 10000, j = 1, 2, ..., n)(1≤kj?≤10000,j=1,2,...,n) — the number of lamps in the rooms of the Castle. The number in position jjj is the number of lamps in jjj-th room. Room numbers are given in accordance with Lpl's list.
The third line contains one integer q(1≤q≤100000)q (1 \le q \le 100000)q(1≤q≤100000) — the number of queries.
The fourth line contains qqq integers d1,d2,...,dqd_1, d_2, ..., d_qd1?,d2?,...,dq?
(1≤dp≤100000,p=1,2,...,q)(1 \le d_p \le 100000, p = 1, 2, ..., q)(1≤dp?≤100000,p=1,2,...,q) — numbers of months, in which queries are formed.
Months are numbered starting with 111; at the beginning of the first month Lpl buys the first m energy-saving lamps.
Output
Print qqq lines.
Line ppp contains two integers — the number of rooms, in which all old lamps are replaced already, and the number of remaining energy-saving lamps by the end of dpd_pdp? month.
Hint
Explanation for the sample:
In the first month, he bought 444 energy-saving lamps and he replaced the first room in his list and remove it. And then he had 111 energy-saving lamps and skipped all rooms next. So, the answer for the first month is 1,1??????11,1------11,1??????1 room's lamps were replaced already, 111 energy-saving lamp remain.
样例输入复制
5 4 3 10 5 2 7 10 5 1 4 8 7 2 3 6 4 7
样例输出复制
4 0 1 1 3 6 5 1 5 1 2 0 3 2 4 4 3 6 5 1
题目来源
ACM-ICPC 2018 南京赛区网络预赛
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通用
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变量
函数
79
ll powmod(ll x,ll y){ll t;for(t=1;y;y>>=1,x=x*x%mod) if(y&1) t=t*x%mod;return t;}
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while(scanf("%d%d",&n,&m)!=EOF)
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{
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build(1,n,1);///顺便写入数组
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int tv=0,cnt=0;///tm表示最后一次修改的位置
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for(int i=1;i<maxn;i++)
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{
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tv+=m;
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while(~query(1,n,1,tv)) cnt++;
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ans1[i]=cnt,ans2[i]=tv;///已经换的次数，剩余的价值
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///cout<<ans1[i]<<" "<<ans2[i]<<endl;
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}
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scanf("%d",&q);
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for(int i=0;i<q;i++)
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{
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scanf("%d",&x);
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printf("%d %d\n",ans1[x],ans2[x]);
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}
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}
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return 0;
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}
79
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#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
using namespace std;#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)#define root l,r,rt
#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define ll long longconst int maxn =1e5+5;
const int mod=1e9+7;
const int UB=maxn*1000;ll gcd(ll x,ll y) { return y==0?x:gcd(y,x%y); }
ll powmod(ll x,ll y){ll t;for(t=1;y;y>>=1,x=x*x%mod) if(y&1) t=t*x%mod;return t;}
/*
题目大意:换灯泡，每个月都会增加m个价值，
然后每个房间都有对应的价值，必须按序扫，如果手中的价值大于等于房间的价值则必须换。
询问给定月数末尾总共换了多少次，手中剩余多少价值。线段树操作秀一下就行。
查询就查询序列中第一个小于等于给定值的位置，二分一下，
如果左边区间的最小值符合要求则查询左边，否则查询右边，
边界时，判定一下是否符合题意，然后更新就行，否则返回-1表示不可行。线段树的基础题。
顺便想提下线段树的数据对应关系，线段树的数据源应该是由树的
叶节点存储，然后节点价值是由rt索引，minv[rt]存储的，是子树中数据源的函数关系。
该函数满足区间性质。
*/int n,m;
///线段树
int minv[maxn<<2];
void pushup(lrt) { minv[rt]=min(minv[rt<<1],minv[rt<<1|1]); }
void build(lrt)
{minv[rt]=UB;///最大值if(l==r) { scanf("%d",&minv[rt]); return ; }int mid=l+r>>1;build(lson),build(rson),pushup(root);
}
int query(lrt,int& v)///整体查询第一个小于等于v的位置
{if(l==r){if(v<minv[rt]) return -1;else{v-=minv[rt];minv[rt]=UB;///这个点就不会再被选中了return rt;}}int mid=l+r>>1,ret;if(minv[rt<<1]<=v) ret=query(lson,v);else ret=query(rson,v);pushup(root);///因为查询时参杂着修改，所以每次都要刷新return ret;
}///
int x,q;
int ans1[maxn],ans2[maxn];int main()
{while(scanf("%d%d",&n,&m)!=EOF){build(1,n,1);///顺便写入数组int tv=0,cnt=0;///tm表示最后一次修改的位置for(int i=1;i<maxn;i++){tv+=m;while(~query(1,n,1,tv)) cnt++;ans1[i]=cnt,ans2[i]=tv;///已经换的次数，剩余的价值///cout<<ans1[i]<<" "<<ans2[i]<<endl;}scanf("%d",&q);for(int i=0;i<q;i++){scanf("%d",&x);printf("%d %d\n",ans1[x],ans2[x]);}}return 0;
}