水题小技巧

Digital Roots

题一：

题目链接：
http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1009&cid=29312
题意：将一个多位数的各个位上的数字求和，直到这个和是一个个位数。

Problem Description

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed
and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process
must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

Input

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

Output

For each integer in the input, output its digital root on a separate line of the output.

Sample Input

24
39
0

Sample Output

6
3

Source

代码：

#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int main()
{
int n,sum;
while(scanf("%d",&n))
{
if(n==0)
break;
while(n>=10)
{
sum=0;
while(n!=0)
{
sum+=n%10;
n/=10;

}
n=sum;
}
cout<<sum<<endl;

}
return 0;
}


题二：

题目链接：
http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1004&cid=29312

Rightmost Digit

题/*

末位数      相乘后的末位数

1           1

2           4   8   6   2

3           9   7   1   3

4           6   4

5           5

6           6

7           9   3   1   7

8           4   2   6   8

9           1   9（由于x9是奇数，所以1不会出现）

0           0意：N^N的末位数字，规律题，最大循环是4

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 67   Accepted Submission(s) : 18

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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<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2
3
4

Sample Output

7
6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.

In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

Author

代码：

#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int main()
{
int t,n,b;
cin>>t;
while(t--)
{
cin>>n;
b=n%10;
int m=1;
for(int i=0;i<=(n-1)%4;i++)///末位数字最大循环是4
{
m*=b;
if(m>10)
m%=10;
}
cout<<m<<endl;
}
return 0;

}


题三：

改革春风吹满地

题意:求多边形的面积



可以先根据逆时针给出的三个顶点求三角形的面积

三点为A（a,b），B（c,d），C（e,f）

面积公式s=(a*d+b*e+c*f-a*f-b*c-d*e);

一重循环即可

题目链接：
http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1005&cid=29312
代码：

#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
double area(int a,int b,int c,int d,int e,int f)
{
return (a*d+b*e+c*f-a*f-b*c-d*e)*0.5;
}
int main()
{
int t;
int x[100],y[100];
while(scanf("%d",&t))
{
if(t==0)
break;

double s=0;
for(int i=0;i<t;i++)
cin>>x[i]>>y[i];
for(int i=3;i<=t;i++)
s+=area(x[0],y[0],x[i-2],y[i-2],x[i-1],y[i-1]);
cout<<s<<endl;
}
return 0;
}