标签： 数学题

Problem Description
HOHO，终于从Speakless手上赢走了所有的糖果，是Gardon吃糖果时有个特殊的癖好，就是不喜欢将一样的糖果放在一起吃，喜欢先吃一种，下一次吃另一种，这样；可是Gardon不知道是否存在一种吃糖果的顺序使得他能把所有糖果都吃完？请你写个程序帮忙计算一下。

Input
第一行有一个整数T，接下来T组数据，每组数据占2行，第一行是一个整数N（0注意大数组要放在main()函数之外，否则RE。sum 必须用_int64开才行，long long int 会导致Runtime Error（OJ傲娇？！）
sort函数逆序排列的方法就是先定义一个比较函数cmp。然后直接在sort中使用~
悲剧的RE了无数次……
AC代码：

#include
#include
#include
using namespace std;
int cdy[1000001];
bool cmp(int a, int b)
{
	return a>b;
}
int main()
{
	int n, i,num;
	_int64 sum;
	scanf("%d", &n);
	while (n--)
	{
		sum = 0;
		scanf("%d", &num);
		for (i = 0; isum+1)
		{
			printf("Non");
		}
		else
		{
			printf("Yesn");
		}
	}
	return 0;
}

Problem Description
Given a positive integer N, you should output the most right digit of N^N.

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1 #include using namespace std; int main(){ int n,m; int a[10][4] = {{0,0,0,0},{1,1,1,1},{2,4,8,6},{3,9,7,1},{4,6,4,6},{5,5,5,5},{6,6,6,6},{7,9,3,1},{8,4,2,6},{9,1,9,1}}; cin>>m; while(m–) cin>>n,cout=0?n%4-1:3]

Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

Input
Input consists of a sequence of lines, each containing an integer n. (n #include using namespace std; int main(){ int n; while(cin>>n){ if(n%4==2) cout

Problem Description
The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input
There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output
Print the total time on a single line for each test case.

Sample Input
1 2
3 2 3 1
0

Sample Output
17
41

Author
ZHENG, Jianqiang

别人问的一道题目，没什么好说的。要注意即便是相同楼层也要停5s才行。

样例：

3 1 1 1

输出：

21

 
代码：

#include
int main()
{
	int n;
	while (~scanf("%d", &n)&&n!=0)
	{
		int f, fi,j, sum;
		f = sum = 0;
		for (j = 0; j f)
			{
				sum += 4 * (fi - f);
			}
			//if (j!=n-1)注意最后也要停五秒
				sum += 5;
		}
		printf("%dn", sum);
	}
	return 0;
}

临近期末，心浮气躁。来点水题降降火~

The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7807 Accepted Submission(s): 5483

Problem Description
Ignatius bought a land last week, but he didn’t know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture, can you tell Ignatius the area of the land?

Note: The point P1 in the picture is the vertex of the parabola.

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).

Output
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

Sample Input
2
5.000000 5.000000
0.000000 0.000000
10.000000 0.000000
10.000000 10.000000
1.000000 1.000000
14.000000 8.222222

Sample Output
33.33
40.69

Hint

For float may be not accurate enough, please use double instead of float.

数学问题，直接上代码：

#include
#include
int main()
{
    int n;
 	double x1,y1,x2,y2,x3,y3;
  	double a,b,c,k,t;
	double area1,area2;
    while(~scanf("%d",&n))
    while(n--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
        a=(y2-y1)/pow((x2-x1),2);
        b=-2*a*x1;
        c=y1-a*x1*x1-b*x1;
        k=(y3-y2)/(x3-x2);
        t=y3-k*x3;
        area1=1.0/3*a*pow(x3,3)+1.0/2*(b-k)*pow(x3,2)+(c-t)*x3;
        area2=1.0/3*a*pow(x2,3)+1.0/2*(b-k)*pow(x2,2)+(c-t)*x2;
        printf("%.2lfn",area1-area2);
    }
    return 0;
}