There are two popular formats for representing a date: day/month/year or month/day/year. For example, today can be represented as 15/8/2017 or 8/15/2017.
Sometimes (like on today), using one way or another should pose no confusion — it is immediately understood that the date is the 15th of August. On other days, however, the two representations may be interpreted as two different valid dates. For example, the 7th of August may be misinterpreted as the 8th of July, since both can be represented as 7/8/2017 (or 8/7/2017).
We say a date (D, M, Y) is ambiguous if D/M/Y and M/D/Y, when both interpreted in the day/month/year format, are different valid dates. For example, (7, 8, 2017) and (8, 7, 2017) are ambiguous, while (15, 8, 2017) and (10, 10, 2017) are not.
The total number of ambiguous dates in the Gregorian calendar system on any given year is equal to 12 × 11 = 132.
Now, suppose that in a hypothetical calendar system, there are M months, where the i-th month has D[i] days, numbered from 1 to D[i]. Assume that there are no leap years.
You are to carry out a calendar reform, by shuffling the array D[], and your target is to minimize the total number of ambiguous dates in a calendar year. Specifically, you want to find a permutation p[1], p[2], ..., p[M] of integers 1, 2, ..., M, such that the new calendar system, where the i-th month has D[p[i]] days, has the minimal number of ambiguous dates. Output that minimal number.
Input
The first line of input consists of a single integer M, the number of months in the hypothetical calendar system.
The second line of input consists of M integers D[1], D[2], ..., D[M], the original number of days in the i-th month.
For all test cases, 1 ≤ M ≤ 105, 1 ≤ D[i] ≤ 105.
Output
Output a single integer, the minimal number of ambiguous dates after the calendar reform.
Example
12
31 28 31 30 31 30 31 31 30 31 30 31
132
3
5 1 1
0
逻辑题，显然某个月份的天数越多，这个月就应该越靠后。
想通就很容易了，排个序就好了。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> #include<queue> using namespace std; const int MAXN=1e5+10; const double mod=1e16+7; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) int main() { int n,a[MAXN]; ll ans=0,sum; scanf("%d",&n); FOR(i,1,n) scanf("%d",&a[i]); sort(a+1,a+n+1); ans+=2*min(a[1]-1,n-1); FOR(i,2,n) { if(a[i]>i) ans+=2*min(a[i]-i,n-i); } printf("%lld ",ans); }
To boost contestants' performances in the 20th La Salle - Pui Ching Programming Challenge, the organizers have bought N robots to cheer for them. Each robot is supposed to display cheering slogans, letter by letter.
Unfortunately, due to some technical reasons, the display screen of each robot can only display one fixed character. Therefore, the organizers decided to arrange the robots in a row, thus forming a string of N letters (What a waste!). All letters are in uppercase.
The two hosting schools have abbreviated names LSC and PCMS, as we all know. Which of the two names appear (as a substring) in the string more often?
Input
The first and only line of input consists of a string with N uppercase letters.
For all test cases, N ≤ 100.
Output
Let A be the number of occurrences of LSC in the given string.
Let B be the number of occurrences of PCMS in the given string.
If A > B, output LSC.
If A < B, output PCMS.
If A = B, output Tie.
签到题，直接暴力就好了，不过我被卡了一下，似乎是我的编译系统和评测系统不同。
举个例子for（int i=1;i<1;i++)在我这不会进行运算，在评测那里就导致访问越界了。
#include<bits/stdc++.h> using namespace std; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define useit ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) int main() { string s; cin>>s; int l1=0,l2=0; if(s.length()>1) FOR(i,2,s.length()-1) { //cout<<s[i-2]<<s[i-1]<<s[i]<<endl; if(s[i-2]=='L'&&s[i-1]=='S'&&s[i]=='C') l1++; } if(s.length()>2) FOR(i,3,s.length()-1) { //cout<<s[i-3]<<s[i-2]<<s[i-1]<<s[i]<<endl; if(s[i-3]=='P'&&s[i-2]=='C'&&s[i-1]=='M'&&s[i]=='S') l2++; } if(l1>l2) cout<<"LSC"<<endl; else if(l1<l2) cout<<"PCMS"<<endl; else cout<<"Tie"<<endl; }
The Gregorian calendar is internationally the most widely used civil calendar. It is named after Pope Gregory XIII, who introduced it in October 1582.
In the Gregorian calendar, there are 28 days in February in a common year and 29 days in February in a leap year. Year Y is a leap year if and only if Y is a multiple of 400, or Y is a multiple of 4 and is not a multiple of 100.
Percy is curious about the distribution of days of the week of his birthday in his life. By checking the calendar, he quickly finds that in the years between 1999 and 2017 (inclusive), his birthday (in case you do not know, 27 February) appears only twice on both Tuesday and Thursday, three times on each of the other days of the week.
Percy finds counting the distribution of some days in some consecutive years really cool, so he decides to invent a way to quickly count the distribution.
Within 15 minutes, he successfully invented a fast program to do the calculation for years between 1583 and 2 × 109, inclusive. His program can answer 5000 queries in 1 second. However, he is not sure if the program works correctly, so he needs your help. Your task is simple, write your own program to do the calculation, so that Percy can check his program's correctness by comparing the outputs of different queries with your program.
In this problem, please assume the definition of leap years mentioned above is true for all years between 1583 and 2 × 109, inclusive.
Input
The first line consists of a single integer, Q, denotes the number of queries. (1 ≤ Q ≤ 5000)
In the next Q lines, each describes a single query. The queries are in the format S E M D, which means you have to calculate the distribution of days of the week for the D-th day of the M-th month for all years between S and E, inclusive. (1583 ≤ S ≤ E ≤ 2 × 109, the days given are one of the 366 valid days)
Output
Output Q lines, each answers a query given.
In each line output 7 integers, the frequencies of days of the weeks in this order: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
The order of answers should follow the order of queries given.
Example
1
1999 2017 2 27
3 3 2 3 2 3 3
2
2017 2017 8 15
2017 2021 2 29
0 0 1 0 0 0 0
0 0 0 0 0 0 1
4
3141 5926 5 3
5897 9323 8 4
2718 2818 2 8
2222 2222 2 22
404 391 403 390 404 396 398
488 488 497 481 497 480 496
15 14 14 15 14 15 14
0 0 0 0 0 1 0
日期模拟，发现400年为一个完整的循环（日期星期闰年状态相同），然后要预处理，建议用一个三维数组存情况，这样子写代码会很轻松。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> #include<queue> using namespace std; const int MAXN=1e5+10; const double mod=1e16+7; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) int isr(int n) { if(n%4==0&&n%100!=0||n%400==0)return 1; else return 0; } int da[810][15][35]; //20871 void init() { int k[15],l=6; k[1]=31,k[2]=28,k[3]=31,k[4]=30,k[5]=31,k[6]=30; k[7]=31,k[8]=31,k[9]=30,k[10]=31,k[11]=30,k[12]=31; mst(da); FOR(i,0,399) { FOR(j,1,12) { if(j==2) FOR(t,1,k[j]+isr(i)) { if(l+1==8)l=0; da[i][j][t]=++l,da[i+400][j][t]=l; } else FOR(t,1,k[j]) { if(l+1==8)l=0; da[i][j][t]=++l,da[i+400][j][t]=l; } } } } int main() { int q,y1,y2,m,d,t; ll s[10]; cin>>q; init(); while(q--) { mst(s); scanf("%d%d%d%d",&y1,&y2,&m,&d); if(y2-y1>=400) { t=(y2-y1)/400; FOR(i,y1%400,y1%400+399) s[da[i][m][d]]+=t; } y1%=400,y2%=400; if(y1>y2) y2+=400; FOR(i,y1,y2) { s[da[i][m][d]]++; } FOR(i,1,7) cout<<s[i]<<" "; edl; } }
"Hit!" is a popular game in ancient Byteland.
The very first version of the game is quite simple: each player picks up a stone and throws it at a circle drawn on the ground. A player wins if his/her stone lands inside the circle.
After 20 years of practice, Bitman, a young man living in ancient Byteland, has mastered the skill of throwing stones – he can throw a stone at any specific place he wants. With such skill, Bitman plays "Hit!" without losing a single game. He simply targets every stone at the center of the circle!
The King of Hackerland hears the story of Bitman and wants to challenge him with a harder, though still very simple, version of "Hit!".
In each game, two circles which share a positive common area are drawn on the ground. In order to win, the player must throw a stone at the common area of the two circles.
As Bitman had no idea how to target his stone at the common area, he asks for your help. Given the coordinates of the centers and radii of the two circles, please tell Bitman the coordinates of any point he can target at such that he can win the game.
For simplicity, you can consider the landing position of the stone as a single point.
Input
The input consists of two lines, each describes one circle drawn on the ground. Each line contains three integers x, y and r, denoting respectively the x-coordinate, y-coordinate, and the radius of a circle.
All coordinates have their absolute value no more than 100, and 1 ≤ r ≤ 100 for both circles.
Output
Output two numbers, the x-coordinate and y-coordinate of a point where Bitman can throw his stone at to win the game.
Your answer will be accepted if for each of the two circles, the point lies inside the circle or that the distance between the point and the circle is not greater than 10 - 5.
Example
0 0 3
3 4 3
1.5 2.5
-7 -9 3
-4 -4 5
-6 -7
Note
In the first sample, (1.5, 2.5) is a possible answer as it lies inside the common area of two circles drawn. Please note that there exists more than one possible answer in this case. For example, (2, 2), (1, 2) and (2.1, 1.87) are also possible answers.
其实题目对于精度没有很高的要求，求出圆心连线与两个圆的交点，再输出这两个点的中点就能满足精度要求，但这题有个坑点，如果连线不与圆相交就会出错，需要特判
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> #include<queue> using namespace std; const int MAXN=1e5+10; const double eps=1e-6; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) struct point { double x,y; point() {} point(double _a,double _b) { x=_a; y=_b; } }; double distan(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } double dist(point p1,point p2) { return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y); } int main() { point a,b,c,d,e; double r1,r2,l,s; cin>>a.x>>a.y>>r1>>b.x>>b.y>>r2; l=distan(a,b); if(l<r1||l<r2) { if(l<r1) cout<<b.x<<" "<<b.y<<endl; else if(l<r2) cout<<a.x<<" "<<a.y<<endl; } else { s=l-r2; c.x=a.x*l-(a.x-b.x)*s; c.y=a.y*l-(a.y-b.y)*s; s=l-r1; d.x=b.x*l-(b.x-a.x)*s; d.y=b.y*l-(b.y-a.y)*s; e.x=(c.x+d.x)/2/l; e.y=(c.y+d.y)/2/l; /*while(dist(a,e)>r1*r1) { e.x+=(b.x-a.x)*eps; } while(dist(b,e)>r2*r2) { e.x-=(b.x-a.x)*eps; }*/ cout<<e.x<<" "<<e.y<<endl; }
Arya lives in a magical world. You can view it as a number line.
There are N citizens in total. The i-th citizen have his own house in position i with height Hi, Noted that Hi could be non positive, which means their house is actually built beneath the horizon.
Arya is the hand of the king and she thinks that the buildings are a bit out of order. She denote the chaos index of the world as .
Noted that hi could be equal to zero as the world is magical.
She feels that the current chaos index of the world is too high. So she designs to change the world a bit by using her super power.
She could flip a continuous sequence of building, i.e., she could choose two arbitrary integer L and R, where 1 ≤ L ≤ R ≤ N, and invert the signs of Hi for all L ≤ i ≤ R (positive to negative and vice versa).
As Arya is weak, she would only do this operation exactly once. Arya wants to minimize the chaos index after the operation. Being a good guy, help Arya to find the lowest possible chaos index after exactly flipping one continuous sequence of building.
Input
The first line contains an integer N. (1 ≤ N ≤ 106)
The second line consist of N integers, the i-th integer is Hi. ( - 109 ≤ Hi ≤ 109)
Output
Output consist only one integer in a single line, the lowest possible chaos index after exactly flipping one continuous sequence of building.
Example
4
1 -2 -3 4
3
5
-3 -2 0 -5 3
10
Note
In the first sample, Arya should flip house 2 to 3, so that the heights become 1, 2, 3, 4 and the chaos index would be |2 - 1| + |3 - 2| + |4 - 3| = 3.
|x - y| is the absolute difference between x and y.
逻辑题，当你意识到选择一个区间只会对两个点产生影响时，答案就很明显了。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> #include<queue> using namespace std; const int MAXN=1e6+10; const double mod=1e16+7; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) int n,a[MAXN],l1=0,l2=0,s1=0,s2=0,k; int main() { ll ans=0; scanf("%d%d",&n,&a[1]); FOR(i,2,n) { scanf("%d",&a[i]); ans+=abs(a[i]-a[i-1]); if(a[i]>0&&a[i-1]<0||a[i]<0&&a[i-1]>0) { if(l1==0) { l1=i; s1=2*min(abs(a[i]),abs(a[i-1])); } else { if(l2==0) { l2=i; s2=2*min(abs(a[i]),abs(a[i-1])); if(s2>s1) swap(l1,l2),swap(s1,s2); } else { k=2*min(abs(a[i]),abs(a[i-1])); if(k>s1) l2=l1,s2=s1,l1=i,s1=k; else if(k>s2) l2=i,s2=k; } } } } printf("%lld ",ans-s1-s2); }
The King of Byteland owns an army of well-trained knights. One day, he decides to attack Hackerland. Hackerland can be described as a grid with N rows and M columns. The cell on the i-th row, j-th column is called cell (i, j) for convenience. The north-west corner is (1, 1) while the south-east corner is (N, M).
In order to win, the Byteland army must occupy all cells with knights. Now, the King of Byteland has carefully planned the attack. The attack can be divided into two phases.
In the first phase, a number of knights are sent to some specific cells of Hackerland.
After the first phase has ended, the second phase begins. In the second phase, The knights will start attacking the land of Hackerland. Their only way to attack is as follows: If at any moment, there are two knights on the same row or the same column, then all unoccupied cells between the two knights will be under attack, and more knights will be immediately sent to occupy these cells. This phase will end when no more cells can be occupied.
Please note that the knights are not allowed to move once they are sent to Hackerland.
In the following graph, the cells (2, 7), (4, 3), (5, 7) and (8, 3) are occupied in the first phase.
When the second phase starts, the cells between (4, 3) and (8, 3), (2, 7) and (5, 7) are under attack, and then occupied by other knights. Afterwards, the cells between (4, 3) and (4, 7), (5, 3) and (5, 7) will also be occupied. The second phase will end after this as no more cells can be occupied.
The attack is currently in the first phase. K cells of Hackerland are already occupied under the command of the King. You, the leader of the knights, are going to send some more knights (possibly none) to some cells of Hackerland of your choice before the second phase starts.
On one hand, you have to send enough knights to ensure that at the end of the war, all cells are occupied. On the other hand, you have to send as few knights as possible to minimize the chance of being discovered by the army of Hackerland. Your task is to find out the minimum number of knights to be sent.
Input
The first line contains three integers: N, M and K. (1 ≤ N, M ≤ 100, 0 ≤ K ≤ N·M)
The next K lines each contains two integers: x and y, which means a knight was sent by the King to occupy cell (x, y). (1 ≤ x ≤ N, 1 ≤ y ≤ M)
No any two knights are sent to the same cell.
Output
Output an integer, the minimum number of extra knights to be sent, on a single line.
Example
1 1 0
1
2 2 0
4
3 3 8
1 1
1 2
1 3
2 1
2 3
3 1
3 2
3 3
0
逻辑题，如果玩过黑白棋就知道四个角是什么地位了，其他的位置不会干扰到角。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> #include<queue> using namespace std; const int MAXN=1e5+10; const double mod=1e16+7; #define INF 0x7fffffff #define ll long long #define edl putchar(' ') #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) int main() { int n,m,k,ans,a,b; scanf("%d%d%d",&n,&m,&k); if(n==1||m==1) { if(n==m)ans=1; else ans=2; if(n==1) FOR(i,1,k) { scanf("%d%d",&a,&b); if(b==1||b==m)ans--; } else if(m==1) FOR(i,1,k) { scanf("%d%d",&a,&b); if(a==1||a==n)ans--; } } else { ans=4; FOR(i,1,k) { scanf("%d%d",&a,&b); if(a==1&&b==m||a==1&&b==1||a==n&&b==1||a==n&&b==m)ans--; } } printf("%d ",ans); }