Rope Folding [Brian Dean, 2012]时间限制: 1 Sec 内存限制: 128 MB
题目描述
Farmer John has a long rope of length L (1 <= L <= 10,000) that he uses for various tasks around his farm. The rope has N knots tied into it at various distinct locations (1 <= N <= 100), including one knot at each of its two endpoints.
FJ notices that there are certain locations at which he can fold the rope back on itself such that the knots on opposite strands all line up exactly with each-other:
Please help FJ count the number of folding points having this property. Folding exactly at a knot is allowed, except folding at one of the endpoints is not allowed, and extra knots on the longer side of a fold are not a problem (that is, knots only need to line up in the areas where there are two strands opposite each-other). FJ only considers making a single fold at a time; he fortunately never makes multiple folds.
输入
* Line 1: Two space-separated integers, N and L.
* Lines 2..1+N: Each line contains an integer in the range 0...L specifying the location of a single knot. Two of these lines will always be 0 and L.
输出
* Line 1: The number of valid folding positions.
样例输入
5 10
0
10
6
2
4
样例输出4
提示INPUT DETAILS:
The rope has length L=10, and there are 5 knots at positions 0, 2, 4, 6, and 10.
OUTPUT DETAILS:
The valid folding positions are 1, 2, 3, and 8.
A题
此题巨坑。。。注意题中并没说折点是整数点。。。
Overplanting (Bronze) [Brian Dean, 2012]时间限制: 1 Sec 内存限制: 128 MB
题目描述Farmer John has purchased a new machine that is capable of planting grass within any rectangular region of his farm that is "axially aligned" (i.e., with vertical and horizontal sides). Unfortunately, the machine malfunctions one day and plants grass in not one, but N (1 <= N <= 10) different rectangular regions, some of which may even overlap.
Given the rectangular regions planted with grass, please help FJ compute the total area in his farm that is now covered with grass.
输入
* Line 1: The integer N.
* Lines 2..1+N: Each line contains four space-separated integers x1 y1 x2 y2 specifying a rectangular region with upper-left corner (x1,y1) and lower-right corner (x2,y2). All coordinates are in the range -10,000...10,000.
输出
* Line 1: The total area covered by grass.
样例输入
2
0 5 4 1
2 4 6 2
样例输出20
B题
这题有几种做法：①切割（队列或递归模拟队列） ②线段树 ③容斥原理
Moo [Brian Dean, 2012]时间限制: 1 Sec 内存限制: 128 MB
题目描述The cows have gotten themselves hooked on a new word game, called "Moo". It is played by a group of cows standing in a long line, where each cow in sequence is responsible for calling out a specific letter as quickly as possible. The first cow who makes a mistake loses.
The sequence of letters in Moo can technically continue forever. It starts like this:
m o o m o o o m o o m o o o o m o o m o o o m o o m o o o o o
The sequence is best described recursively: let S(0) be the 3-character sequence "m o o". Then a longer sequence S(k) is obtained by taking a copy of the sequence S(k-1), then "m o ... o" with k+2 o‘s, and then another copy of the sequence S(k-1). For example:
S(0) = "m o o"
S(1) = "m o o m o o o m o o"
S(2) = "m o o m o o o m o o m o o o o m o o m o o o m o o"As you can see, this process ultimately builds an infinitely long string, and this is the string of characters used for the game of Moo.
Bessie the cow, feeling clever, wishes to predict whether the Nth character of this string will be an "m" or an "o". Please help her out!
输入
* Line 1: A single integer N (1 <= N <= 10^9).
输出
* Line 1: The only line of output should contain a single character, which is either m or o.
样例输入
11
样例输出m
提示INPUT DETAILS:
Bessie wants to predict the 11th character.
C题
水题，递归N在这个序列中的位置即可，记f(i) = size(s(i)) , 显然f(i) = 2*f(i-1) + n +
3 ， 可以推下它的通项可以发现 i<=32，找到n>=f(i)的最大i值，递归即可。
Work Scheduling [Richard Peng, 2008]时间限制: 1 Sec 内存限制: 128 MB
题目描述
Farmer John has so very many jobs to do! In order to run the farm efficiently, he must make money on the jobs he does, each one of which takes just one time unit.
His work day starts at time 0 and has 1,000,000,000 time units (!). He currently can choose from any of N (1 <= N <= 100,000) jobs conveniently numbered 1..N for work to do. It is possible but extremely unlikely that he has time for all N jobs since he can only work on one job during any time unit and the deadlines tend to fall so that he can not perform all the tasks.
Job i has deadline D_i (1 <= D_i <= 1,000,000,000). If he finishes job i by then, he makes a profit of P_i (1 <= P_i <= 1,000,000,000).
What is the maximum total profit that FJ can earn from a given list of jobs and deadlines? The answer might not fit into a 32-bit integer.
输入
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: D_i and P_i
输出
* Line 1: A single number on a line by itself that is the maximum possible profit FJ can earn.
样例输入
3
2 10
1 5
1 7
样例输出17
提示OUTPUT DETAILS:
Complete job 3 (1,7) at time 1 and complete job 1 (2,10) at time 2 to maximize the earnings (7 + 10 -> 17).
Gold B题
仔细分析不难发现对于前i时间，我们只需保留i个价值最大的任务，那么可以用堆维护。
1 #include<cstdio>
2 #include<cstdlib>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 #define For(i,n) for(int i=1;i<=n;i++)
7 int ans,A[10100],n,m,x;
8
9 int main(){
10 scanf("%d%d",&n,&m);
11 For(i,n){
12 scanf("%d",&x);
13 A[x] = 1;
14 }
15 For(i,m){
16 int t = 0 , flag = 0;
17 if(i!=m)
18 while(i>=t&&(i+t<=m)){
19 if(A[i-t]!=A[i+t]) {
20 flag = 1;
21 break;
22 }
23 t++;
24 }
25 else flag = 1;
26 if(!flag) ans++;
27 flag = 0; t = 0;
28 while(i>=t+1&&i+t<=m){
29 if(A[i-t-1]!=A[i+t]){
30 flag = 1;
31 break;
32 }
33 t++;
34 }
35 if(!flag) ans++;
36 }
37 printf("%d\n",ans);
38 return 0;
39 }
40 -----------------------------------以上是A题----------------------------------
41
42 #include<cstdio>
43 #include<cstdlib>
44 #include<cstring>
45 #include<iostream>
46 #include<algorithm>
47 using namespace std;
48 const int NN = 101;
49 #define For(i,n) for(int i=1;i<=n;i++)
50 #define Rep(i,l,r) for(int i=l;i<=r;i++)
51 struct Rectangle{
52 int bx,by,tx,ty;
53 }R[NN],Temp;
54
55 int n;
56 int x1,x2,y1,y2;
57 long long ans;
58
59 int GetArea(int X1,int Y1,int X2,int Y2){
60 return abs(X2-X1)*abs(Y2-Y1);
61 }
62
63 int Cover(int lbx,int lby,int ltx,int lty,int Height){
64 int ANS=0;
65 if(Height==n) return GetArea(lbx,lby,ltx,lty);
66 int tbx=R[Height+1].bx,tby=R[Height+1].by,ttx=R[Height+1].tx,tty=R[Height+1].ty;
67 if(lby<tby&<x>tbx&&ttx>lbx)//左边
68 ANS+=Cover(max(tbx,lbx),lby,min(ltx,ttx),min(lty,tby),Height+1);
69 if(lty>tty&<x>tbx&&ttx>lbx)//右边
70 ANS+=Cover(max(tbx,lbx),max(tty,lby),min(ttx,ltx),lty,Height+1);
71 if(lbx<tbx)//下边
72 ANS+=Cover(lbx,lby,min(ltx,tbx),lty,Height+1);
73 if(ltx>ttx)//上边
74 ANS+=Cover(max(lbx,ttx),lby,ltx,lty,Height+1);
75 return ANS;
76 }
77
78 int main(){
79 scanf("%d",&n);
80 For(i,n){
81 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
82 swap(x1,y1);swap(x2,y2);
83 R[i].bx = min(x1,x2);
84 R[i].by = min(y1,y2);
85 R[i].tx = max(x1,x2);
86 R[i].ty = max(y1,y2);
87 }
88 for(int i=n;i>=1;i--)
89 ans+=Cover(R[i].bx,R[i].by,R[i].tx,R[i].ty,i);
90 cout<<ans<<endl;
91 return 0;
92 }
93 ----------------------------------以上是B题-----------------------------------
94 #include<cstdio>
95 #include<cstdlib>
96 #include<cstring>
97 #include<iostream>
98 #include<algorithm>
99 using namespace std;
100 #define For(i,n) for(int i=1;i<=n;i++)
101 #define Rep(i,l,r) for(int i=l;i<=r;i++)
102 char ANS[5] = {‘m‘,‘m‘,‘o‘,‘o‘};
103 long long f[33];
104 int n;
105
106 void init(){
107 f[0] = 3;
108 For(i,32) f[i] = f[i-1]*2 + i + 3;
109 scanf("%d",&n);
110 }
111
112 char Solve(int K){
113 int sn;
114 if(K<=3) return ANS[K];
115 Rep(i,0,32)
116 if(K<f[i]){
117 sn = i-1;
118 break;
119 }
120 if(K-f[sn]<=sn+4){
121 if(K-f[sn]==1) return ‘m‘;
122 else return ‘o‘;
123 }else
124 return Solve(K-f[sn]-sn-4);
125 }
126
127
128 int main(){
129 init();
130 cout<<Solve(n)<<endl;
131 return 0;
132 }
133 --------------------------------以上是C题-------------------------------------
134 #include<queue>
135 #include<cmath>
136 #include<cstdio>
137 #include<cstdlib>
138 #include<cstring>
139 #include<iostream>
140 #include<algorithm>
141 using namespace std;
142 #define For(i,n) for(int i=1;i<=n;i++)
143 #define Rep(i,l,r) for(int i=l;i<=r;i++)
144
145 struct jobs{
146 int times,profit;
147 }G[100100];
148
149 struct cmp{
150 bool operator () (jobs A,jobs B){
151 return A.profit>B.profit;
152 }
153 };
154
155 priority_queue<jobs,vector<jobs>,cmp> T;
156
157 long long ans;
158 int n,x,y,limits;
159
160 bool CMP(jobs A,jobs B){
161 return A.times<B.times;
162 }
163
164 int main(){
165 scanf("%d",&n);
166 For(i,n){
167 scanf("%d%d",&x,&y);
168 G[i].times = x; G[i].profit = y;
169 }
170 sort(G+1,G+n+1,CMP);
171 For(i,n){
172 T.push(G[i]);
173 limits = G[i].times;
174 while(T.size()>limits) T.pop();
175 }
176 while(T.size()){
177 ans+=T.top().profit;
178 T.pop();
179 }
180 cout<<ans<<endl;
181 return 0;
182 }
183 --------------------------------以上是Gold·B题-------------------------------
ABC&GoldB题代码
USACO·2012·Feb Bronze && 2009·Open Gold