本文介绍了三道算法题目——“Hearthstone”卡牌分配、“Bilibili”弹幕发送及“Triangle”三角形分割的解题思路与AC代码实现。通过数学推导和贪心算法解决实际问题。
Link：http://acm.fafu.edu.cn/problemList.php?volume=6
Problem1557~1572
Hearthstone
| Time Limit: | 1000MS | Memory Limit: | 65536KB |
| Total Submissions: | 63 | Accepted: | 28 |
Cdfpysw loves playing a card game called "Hearthstone".
Now he has N cards, he wants to split these cards into 4 piles. Let's assume the number of cards in each pile is a1, a2, a3, a4.
It must be satisfied that:
a1 * k1 = a2 + a3 + a4
a2 * k2 = a1 + a3 + a4
a3 * k3 = a1 + a2 + a4
a1, a2, a3, a4 must be positive
Because Cdfpysw is clever, there must be a way to split there card. Can you tell Cdfpysw what is the way?
Now he has N cards, he wants to split these cards into 4 piles. Let's assume the number of cards in each pile is a1, a2, a3, a4.
It must be satisfied that:
a1 * k1 = a2 + a3 + a4
a2 * k2 = a1 + a3 + a4
a3 * k3 = a1 + a2 + a4
a1, a2, a3, a4 must be positive
Because Cdfpysw is clever, there must be a way to split there card. Can you tell Cdfpysw what is the way?
The first line is an integer T, means the number of cases.
Then T lines, each line contains 4 integers, N, k1, k2, k3, meaning of these have been shown in the description.
T <= 100
1 <= N <= 10^9
1 <= k1, k2, k3 <= 200
Then T lines, each line contains 4 integers, N, k1, k2, k3, meaning of these have been shown in the description.
T <= 100
1 <= N <= 10^9
1 <= k1, k2, k3 <= 200
For each case, you should output "Case #i: " first, i is the case number.
Then output 4 integers a1, a2, a3, a4, represent the number of cards in each pile.
Then output 4 integers a1, a2, a3, a4, represent the number of cards in each pile.
1120 2 3 4
Case #1: 40 30 24 26
解以下方程组：
a1 * k1 = a2 + a3 + a4 .............(1)
a2 * k2 = a1 + a3 + a4 .............(2)
a3 * k3 = a1 + a2 + a4 .............(3)
a1 + a2 + a3 + a4 = n .............(4)
将（4）式变形后分别代入（1）（2）（3）式子可得：
a1=n/(k1+1);
a2=n/(k2+1);
a3=n/(k3+1);
a4=n-a1-a2-a3;
然后代入输入的数据即可。
AC code：
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
int a1,a2,a3,a4,k1,k2,k3,n,T,cas;
scanf("%d",&T);
for(cas=1;cas<=T;cas++)
{
scanf("%d%d%d%d",&n,&k1,&k2,&k3);
a1=n/(k1+1);
a2=n/(k2+1);
a3=n/(k3+1);
a4=n-a1-a2-a3;
printf("Case #%d: %d %d %d %d\n",cas,a1,a2,a3,a4);
}
return 0;
}
Bilibili
| Time Limit: | 1000MS | Memory Limit: | 65536KB |
| Total Submissions: | 102 | Accepted: | 23 |
Cdfpysw loves to watch videos at bilibili.com.
When he watch videos, he can write some comments, and they will be shown on his screen.
His screen has size of n * m, and every word has size of 1 * 1. When Cdfpysw send a comment, it'll be shown on the rightmost of the topmost line which doesn't contain any comment. If there are comments in all lines, this comment will be failed to send. Every second, the comments on the screen will be shifted to left for 1 before the new comments be send.
e.g.:
Now, the following picture is the screen. Cdfpysw wants to write 3 comments at now, 1 second later, and 6 seconds later. They will be shown on the 2nd line, 5th line, and 1st line.
This example is NOT same as the sample below!
Now Cdfpysw wants to know his comments would be shown on which line, if the comment is failed to send, print "Failed!".
When he watch videos, he can write some comments, and they will be shown on his screen.
His screen has size of n * m, and every word has size of 1 * 1. When Cdfpysw send a comment, it'll be shown on the rightmost of the topmost line which doesn't contain any comment. If there are comments in all lines, this comment will be failed to send. Every second, the comments on the screen will be shifted to left for 1 before the new comments be send.
e.g.:
Now, the following picture is the screen. Cdfpysw wants to write 3 comments at now, 1 second later, and 6 seconds later. They will be shown on the 2nd line, 5th line, and 1st line.
This example is NOT same as the sample below!
Now Cdfpysw wants to know his comments would be shown on which line, if the comment is failed to send, print "Failed!".
The first line is an integer T, means the number of cases.
For each case
The first line contains 3 integers, n, m, q, means the size of screen and the number of comments Cdfpysw wants to send.
Then q lines, for each line, there are 2 integers t, l, means the sending time and the length of the comment. It is guaranteed that t is in ascending order.
T <= 10
1 <= n, m, q <= 50000
0 <= t <= 10^9
1 <= l <= 10^9
For each case
The first line contains 3 integers, n, m, q, means the size of screen and the number of comments Cdfpysw wants to send.
Then q lines, for each line, there are 2 integers t, l, means the sending time and the length of the comment. It is guaranteed that t is in ascending order.
T <= 10
1 <= n, m, q <= 50000
0 <= t <= 10^9
1 <= l <= 10^9
For each case, you should print "Case #i:" in a line, i is the case number.
Then q lines, each line contains an integer to represent the line to show the comment or a string "Failed!" to represent the comment is failed to send.
Then q lines, each line contains an integer to represent the line to show the comment or a string "Failed!" to represent the comment is failed to send.
15 5 70 34 37 58 19 1010 511 5
Case #1:123145Failed!`
编程思想：贪心。比赛时直接暴力模拟就过了，但赛后提交却TLE，必须采用STL的set集合实现nlog（n）的时间复杂度才能过。。
TLE code：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
int row[50050];
int t1,t,n,m,q,l;
int main()
{
//freopen("D:\in.txt","r",stdin);
int T,cas,i,fg,ans;
scanf("%d",&T);
cas=0;
while(T--)
{
cas++;
printf("Case #%d:\n",cas);
scanf("%d%d%d",&n,&m,&q);
memset(row,0,sizeof(row));
t1=0;
while(q--)
{
scanf("%d%d",&t,&l);
l+=m;
fg=0;
for(i=1;i<=n;i++)
{
row[i]-=(t-t1);
if(row[i]<=0)
{
row[i]=0;
if(!fg)
{
fg=i;
printf("%d\n",fg);
row[i]=l;
}
}
}
t1=t;
if(!fg) printf("Failed!\n");
}
}
return 0;
} AC code：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#define LL long long
using namespace std;
int row[50050];
struct node{
int finish;
int id;
bool operator < (const node &a)const{
if(finish!=a.finish) return finish<a.finish;
else return id<a.id;
}
};
set<node>s;
set<node>::iterator it;
set<int>ss;
set<int>::iterator it2;
node p1;
int t,n,m,q,l,cnt;
int main()
{
//freopen("D:\in.txt","r",stdin);
int T,cas,i;
scanf("%d",&T);
cas=0;
while(T--)
{
cas++;
printf("Case #%d:\n",cas);
scanf("%d%d%d",&n,&m,&q);
cnt=0;
s.clear();
ss.clear();
while(q--)
{
scanf("%d%d",&t,&l);
it=s.begin();
while(it!=s.end())
{
if((*it).finish<=t)
{
ss.insert((*it).id);
set<node>::iterator it1=it;
s.erase(it1);
it++;
}
else break;
}
if(ss.empty())
{
++cnt;
if(cnt>n)
printf("Failed!\n");
else
{
printf("%d\n",cnt);
p1.finish=t+l+m;
p1.id=cnt;
s.insert(p1);
}
}
else
{
it2=ss.begin();
printf("%d\n",*it2);
p1.finish=t+l+m;
p1.id=*it2;
s.insert(p1);
ss.erase(*it2);
}
}
}
return 0;
}
Triangle
| Time Limit: | 1000MS | Memory Limit: | 65536KB |
| Total Submissions: | 115 | Accepted: | 36 |
This is a simple question, give you a triangle, please divide the triangle into two parts with a segment.
These two parts should have the same area, and the segment should be as short as possible. Please calculate the length of the segment.
These two parts should have the same area, and the segment should be as short as possible. Please calculate the length of the segment.
The first line of the input is the number T(T<=100), which is the number of cases followed.
For each test case, the only line contains space-separated integers x1, y1, x2, y2, x3, y3,the coordinates of triangle endpoint.
The absolute value of all the coordinates will not be more than 10^9.
For each test case, the only line contains space-separated integers x1, y1, x2, y2, x3, y3,the coordinates of triangle endpoint.
The absolute value of all the coordinates will not be more than 10^9.
For each test case, output only one line, containing the length of segment only. Your answer should be rounded to 4 digits after the decimal point
10 0 10 0 0 10
6.4359
编程思想：推公式。所求的最短边将原三角形最长和次长边截成一个面积为原三角形一半的等腰三角形的两条腰，剩下的就是利用三角形面积公式和余弦定理推出公式来。详见AC代码。
AC code：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#define LL long long
using namespace std;
double dis(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
double a,b,c,x1,y1,x2,y2,x3,y3,ans;
//freopen("D:\in.txt","r",stdin);
int T,cas,i;
scanf("%d",&T);
while(T--)
{
scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
a=dis(x1,y1,x2,y2);
b=dis(x2,y2,x3,y3);
c=dis(x3,y3,x1,y1);
if(b<c) swap(b,c);//使c为三角形的最短边
if(a<c) swap(a,c);
ans=sqrt(a*b-0.5*(a*a+b*b-c*c));
printf("%.4f\n",ans);
}
return 0;
}
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