OUC_Summer Training_ DIV2_#14 724
又落下好多题解啊。。。先把今天的写上好了。
Description
There is a straight snowy road, divided into n blocks. The blocks are numbered from 1 to n from left to right. If one moves from the i-th block to the (i + 1)-th block, he will leave a right footprint on the i-th block. Similarly, if one moves from the i-th block to the (i - 1)-th block, he will leave a left footprint on the i-th block. If there already is a footprint on the i-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the s-th block, makes a sequence of moves and ends in the t-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of s, t by looking at the footprints.
Input
The first line of the input contains integer n (3 ≤ n ≤ 1000).
The second line contains the description of the road — the string that consists of n characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Output
Print two space-separated integers — the values of s and t. If there are several possible solutions you can print any of them.
Sample Input
9
..RRLL...
3 4
11
.RRRLLLLL..
7 5
Hint
The first test sample is the one in the picture.
题目很简单啦，就是给一串足迹，输出一组可能的起始终止的坐标。值得注意的是，本题只有3种情况，全R 全L 有R有L 在第三种情况里只有R在（且全在）L右边的情况。（只有向右走后才有向左走的权利）只输出一种结果就好了。
#include<stdio.h> int main() { char s[1010]; int n,i,a,b; int flag = 0,m = 0,num = 0; scanf("%d",&n); char c = getchar();//直接getchar的话出了错误source code error，但是又交了一次却对了 scanf("%c",&s[0]); for(i = 1;i < n;i++) { scanf("%c",&s[i]); if(s[i] != '.') { num ++; if(m == 0)m = i; if(s[i - 1] != '.'&&flag == 0 && s[i] != s[i - 1])flag = 1; } } if(flag == 0) { if(s[m] == 'R') { a = m + 1; b = a + num; } if(s[m] == 'L') { b = m; a = b + num; } } else { for(i = m;s[i] != '.' && i < n;i++) { if(s[i] == 'L') { a = m+1; b = i; break; } } } printf("%d %d\n",a,b); return 0; }
Description
The polar bears are going fishing. They plan to sail from (sx, sy) to (ex, ey). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (x, y).
- If the wind blows to the east, the boat will move to (x + 1, y).
- If the wind blows to the south, the boat will move to (x, y - 1).
- If the wind blows to the west, the boat will move to (x - 1, y).
- If the wind blows to the north, the boat will move to (x, y + 1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (x, y). Given the wind direction for t seconds, what is the earliest time they sail to (ex, ey)?
Input
The first line contains five integers t, sx, sy, ex, ey (1 ≤ t ≤ 105, - 109 ≤ sx, sy, ex, ey ≤ 109). The starting location and the ending location will be different.
The second line contains t characters, the i-th character is the wind blowing direction at the i-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
Output
If they can reach (ex, ey) within t seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
Sample Input
5 0 0 1 1
SESNW
4
10 5 3 3 6
NENSWESNEE
-1
Hint
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination.
又是一道很简单的题呀，可是却一直wa啊，原来是一个比较简单的地方疏忽了。题的意思是船的行走有四个方向，每次走1（可以不走），给出起点和终点，及每秒的风向。输出到达目的地的最少时间，若不能到达，输出-1.我的方法是，因为船只能走折线到达目的地，根据起末位置确定是N（S）W（E）需要出现的次数即可。
#include<stdio.h> int main() { int t,sx,sy,ex,ey; char str[100010]; scanf("%d%d%d%d%d",&t,&sx,&sy,&ex,&ey); scanf("%s",str); int i,n = 0,s = 0,e = 0,w = 0; if(ex - sx < 0)w = sx - ex; else e = ex - sx; if(ey - sy < 0)s = sy - ey; else n = ey - sy; int N = 0,S = 0,E = 0,W = 0; if((s == 0 && n == 0) &&( w == 0 && e == 0)) { printf("0\n"); return 0; } for(i = 0;i< t;i++) { if(str[i] == 'N'&& n != 0 && N < n)N++;//一开始就是因为没有N < n的条件就一直错。原因是N的次数已经达到条件时若其他条件还未完成N就还会增加，这样肯定就没有结果了 else if(str[i] == 'S' && s!= 0 && S < s)S++; else if(str[i] == 'E' && e != 0 && E < e)E++; else if(str[i] == 'W'&& w != 0 && W < w)W++; if((S == s && W == w)&&( E == e&& N == n)){printf("%d\n",i+1);return 0;} } printf("-1\n"); return 0; }
Description
It is known that there are k fish species in the polar ocean, numbered from 1 to k. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the i-th type of fish be wi, then 0 < w1 ≤ w2 ≤ ... ≤ wk holds.
Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights wi (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 105, 1 ≤ k ≤ 109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.
The second line contains n integers each from 1 to k, the list of fish type caught by Alice. The third line contains m integers each from 1 to k, the list of fish type caught by Bob.
Note that one may have caught more than one fish for a same species.
Output
Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.
Sample Input
3 3 3
2 2 2
1 1 3
YES
4 7 9
5 2 7 3
3 5 2 7 3 8 7
NO
Hint
In the first sample, if w1 = 1, w2 = 2, w3 = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.
In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.
题意是有很多不同种类的鱼，种类用实数代表，题目给出的是鱼的种类数，（质量可以任意定，符合规则就好）相同种类的鱼质量相同，不同种类的鱼随着种类数的递增，质量也递增（相同也行）给出A，B两人所捕的鱼的情况，判断是否可能始A的鱼的总重量 绝对 大于B的鱼的总重量。
我的思路是这样的，若A的鱼的个数大于B的话一定可能（YES），因为可以始所有鱼的重量相同。若A鱼的个数不必B多的话，使A的总重量大于B就有很多种情况了，讨论起来太不方便。但反过来想，使B的总重量小于A的情况只有一种，那就是A中所有鱼的种类都小于或等于B的种类（换句话说，就是对于A中的所有种类，总能在B中找到比A种类大的）。
#include <algorithm> #include <iostream> using namespace std; int a[100010],b[100010]; int main() { int n,m,k; int j,i; int flag = 0; cin>>n>>m>>k; for( i=0;i<n;i++) cin >> a[i]; for( i=0;i<m;i++) cin >> b[i]; sort(a,a+n); sort(b,b+m); if(n > m)cout << "YES\n"; else { for(i = 0,j = 0;i<n && j < m;) { if(a[i] <= b[j]) { i++; j++; } else j++; } if(i==n ) cout<<"NO\n"; else cout << "YES\n"; } return 0; }
