国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學(xué)院 > 開發(fā)設(shè)計(jì) > 正文

【Codeforces 733 C. Epidemic in Monstropolis】+ 模擬

2019-11-14 10:36:10
字體:
供稿:網(wǎng)友

C. Epidemic in Monstropolis time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.

Soon, monsters became hungry and began to eat each other.

One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly. There are no monsters which are being eaten at the same moment. After the monster A eats the monster B, the weight of the monster A increases by the weight of the eaten monster B. In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters in the queue, the i-th of which had weight ai.

For example, if weights are [1,?2,?2,?2,?1,?2] (in order of queue, monsters are numbered from 1 to 6 from left to right) then some of the options are:

the first monster can't eat the second monster because a1?=?1 is not greater than a2?=?2;the second monster can't eat the third monster because a2?=?2 is not greater than a3?=?2;the second monster can't eat the fifth monster because they are not neighbors;the second monster can eat the first monster, the queue will be transformed to [3,?2,?2,?1,?2].

After some time, someone said a good joke and all monsters recovered. At that moment there were k (k?≤?n) monsters in the queue, the j-th of which had weight bj. Both sequences (a and b) contain the weights of the monsters in the order from the first to the last.

You are required to PRovide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn’t make any appointments while monsters were eating each other. Input

The first line contains single integer n (1?≤?n?≤?500) — the number of monsters in the initial queue.

The second line contains n integers a1,?a2,?…,?an (1?≤?ai?≤?106) — the initial weights of the monsters.

The third line contains single integer k (1?≤?k?≤?n) — the number of monsters in the queue after the joke.

The fourth line contains k integers b1,?b2,?…,?bk (1?≤?bj?≤?5·108) — the weights of the monsters after the joke.

Monsters are listed in the order from the beginning of the queue to the end. Output

In case if no actions could lead to the final queue, print “NO” (without quotes) in the only line.

Otherwise print “YES” (without quotes) in the first line. In the next n?-?k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, separated by space, the symbol ‘L’ if the monster which stays the x-th in the queue eats the monster in front of him, or ‘R’ if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again.

When one monster eats another the queue decreases. If there are several answers, print any of them. Examples Input

6 1 2 2 2 1 2 2 5 5

Output

YES 2 L 1 R 4 L 3 L

Input

5 1 2 3 4 5 1 15

Output

YES 5 L 4 L 3 L 2 L

Input

5 1 1 1 3 3 3 2 1 6

Output

NO

Note

In the first example, initially there were n?=?6 monsters, their weights are [1,?2,?2,?2,?1,?2] (in order of queue from the first monster to the last monster). The final queue should be [5,?5]. The following sequence of eatings leads to the final queue:

the second monster eats the monster to the left (i.e. the first monster), queue becomes [3,?2,?2,?1,?2];the first monster (note, it was the second on the previous step) eats the monster to the right (i.e. the second monster), queue becomes [5,?2,?1,?2];the fourth monster eats the mosnter to the left (i.e. the third monster), queue becomes [5,?2,?3];the finally, the third monster eats the monster to the left (i.e. the second monster), queue becomes [5,?5].

Note that for each step the output contains numbers of the monsters in their current order in the queue.

模擬,在a[i]里查詢 == b[i]的數(shù)組區(qū)間l,r,判斷該段數(shù)組里的最大值能否左右去吃,用queue<pair < int,int > > q,記錄過程,

AC代碼:

#include<cstdio>#include<queue>#include<algorithm>using namespace std;queue <pair <int,int> > q;int a[510],b[510],nl,l;bool bc(int L,int R){ if(L == R) return true; int pl = L; for(int i = L; i <= R ; i++) if(a[pl] < a[i]) pl = i; if(pl > L && a[pl] > a[pl - 1]){ for(int i = pl - 1; i >= L ; i--) q.push(make_pair(i - L + nl + 1,'L')); for(int i = pl + 1; i <= R ; i++) q.push(make_pair(nl,'R')); return true; } while(a[pl] == a[pl + 1] && pl < R) pl++; if(pl < R && a[pl] > a[pl + 1]){ for(int i = pl + 1 ; i <= R; i++) q.push(make_pair(pl - L + nl,'R')); for(int i = pl - 1 ; i >= L ; i--) q.push(make_pair(i - L + nl + 1,'L')); return true; } return false;}int main(){ int sum = 0,N,M,ans = 0,t = 1; scanf("%d",&N); for(int i = 1 ; i <= N; i++) scanf("%d",&a[i]),sum += a[i]; scanf("%d",&M); for(int i = 1 ; i <= M; i++) scanf("%d",&b[i]),sum -= b[i]; l = 1,nl = 1; for(int i = 1 ; i <= N; i++){ ans += a[i]; if(ans == b[t]){ ans = 0; if(bc(l,i)) l = i + 1,nl++,t++;//printf("%d/n",l); else{ printf("NO/n");return 0; } } } if(!sum && t == M + 1){ printf("YES/n"); while(!q.empty()) printf("%d %c/n",q.front().first,q.front().second),q.pop(); } else printf("NO/n"); return 0;}
發(fā)表評(píng)論 共有條評(píng)論
用戶名: 密碼:
驗(yàn)證碼: 匿名發(fā)表
主站蜘蛛池模板: 竹溪县| 米泉市| 辽阳县| 昌邑市| 仁化县| 祁连县| 荣成市| 阿鲁科尔沁旗| 友谊县| 合水县| 迭部县| 香格里拉县| 永川市| 眉山市| 奉贤区| 比如县| 长岛县| 纳雍县| 赤峰市| 佛山市| 连山| 惠安县| 长海县| 仁寿县| 左贡县| 永寿县| 罗城| 昭通市| 会理县| 志丹县| 临沂市| 巨鹿县| 汉源县| 磐石市| 安仁县| 高尔夫| 北川| 丹寨县| 古蔺县| 屯昌县| 施甸县|