poj__3145(配套总结本)

Harmony Forever

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 5834 Accepted: 1414

Description

We believe that every inhabitant of this universe eventually will find a way to live together in harmony and peace; that trust, patience, kindness and loyalty will exist between every living being of this earth; people will find a way to appreciate and cooperate with each other instead of continuous bickering, arguing and fighting. Harmony — the stage of society so many people dream of and yet it seems so far away from now…

Fortunately, the method of unlocking the key to true Harmony is just discovered by a group of philosophers. It is recorded on a strange meteorite which has just hit the earth. You need to decipher the true meaning behind those seemingly random symbols… More precisely, you are to write a program which will support the following two kinds of operation on an initially empty setS:

1. B X: Add number X to set S. The K^th command in the form ofB X always happens at time K, and number X does not belong to setS before this operation.
2. A Y : Of all the numbers in set S currently, find the one which has the minimum remainder when divided byY. In case a tie occurs, you should choose the one which appeared latest in the input. Report the time when this element is inserted.

It is said that if the answer can be given in the minimum possible time, true Harmony can be achieved by human races. You task is to write a program to help us.

Input

There are multiple test cases in the input file. Each test case starts with one integerT where 1 ≤ T ≤ 40000. The following T lines each describe an operation, either in the form of “B X” or “A Y ” where 1 ≤X ≤ 500 000, 1 ≤ Y ≤ 1 000 000.

T = 0 indicates the end of input file and should not be processed by your program.

Output

Print the result of each test case in the format as indicated in the sample output. For every line in the form of “A Y”, you should output one number, the requested number, on a new line; output-1 if no such number can be found. Separate the results of two successive inputs with one single blank line.

Sample Input

5B 1A 5B 10A 5A 402B 1A 20

Sample Output

Case 1:121Case 2:1

Source

Shanghai 2006

 

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define LL long long#define pb push_back#define Set(a, v) memset(a, v, sizeof(a))#define For(i, a, b) for(int i = (a); i <= (int)(b); i++)#define Forr(i, a, b) for(int i = (a); i >= (int)(b); i--)#define MAXT (40000+5)#define MAXN (500000+5)#define INF 0x3f3f3f3fbool read(int &x){char ch = getchar();while(ch < '0' || ch > '9') ch = getchar();x = 0;while(ch >= '0' && ch <= '9'){x = x*10+ch-'0';ch = getchar();}return true;}int ql, qr;int id[MAXN];struct Seg_tree{int minv[MAXN*4];void init(){Set(minv, INF); Set(id, -1);}void insert(int o, int L, int R, int v){if(L == R){minv[o] = v; return;}int lc = o<<1, rc = o<<1|1;int mid = (L+R)>>1;if(minv[o] > v) minv[o] = v;if(v <= mid) insert(lc, L, mid, v);else insert(rc, mid+1, R, v);}int query(int o, int L, int R){        if(minv[o] == INF) return INF;if(ql <= L && qr >= R) return minv[o];int lc = o<<1, rc = o<<1|1;int mid = (L+R)>>1;int ret = INF;if(ql <= mid) ret = min(ret, query(lc, L, mid));if(qr > mid) ret = min(ret, query(rc, mid+1, R));return ret;}}ST;char op[5];int nn, num[MAXT];int main(){int n, T = 0, N = 500000;while(read(n) && n){ST.init(); nn = 0;if(T) printf("\n");printf("Case %d:\n", ++T);int x, minv, now, ans, L, R, tot = 0;For(ca, 1, n){scanf("%s", op); read(x);if(op[0] == 'B') ST.insert(1, 0, N, x), num[++nn] = x, id[x] = ++tot;else{if(x <= 5000){minv = INF, ans = -1;Forr(j, nn, 1){if(minv > (num[j]%x)) minv = num[j]%x, ans = j;if(!minv) break;}printf("%d\n", ans);}else{    minv = INF, ans = -1, L = 0;while(L <= N){R = L+x-1;if(R > N) R = N;ql = L; qr = R;now = ST.query(1, 0, N);                        if(now < INF && minv > (now%x)) minv = now%x, ans = id[now];else if(now < INF && minv == (now%x)) minv = now%x, ans = max(ans, id[now]);L = R+1;}printf("%d\n", ans);    }}}}return 0;}