Gold Coins
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 20802 | Accepted: 13018 |
Description
The king pays his loyal knight in gold coins. On
the first day of his service, the knight receives one gold coin. On each of the
next two days (the second and third days of service), the knight receives two
gold coins. On each of the next three days (the fourth, fifth, and sixth days of
service), the knight receives three gold coins. On each of the next four days
(the seventh, eighth, ninth, and tenth days of service), the knight receives
four gold coins. This pattern of payments will continue indefinitely: after
receiving N gold coins on each of N consecutive days, the knight will receive
N+1 gold coins on each of the next N+1 consecutive days, where N is any positive
integer.
Your program will determine the total number of gold coins paid
to the knight in any given number of days (starting from Day 1).
Input
The input contains at least one, but no more than
21 lines. Each line of the input file (except the last one) contains data for
one test case of the problem, consisting of exactly one integer (in the range
1..10000), representing the number of days. The end of the input is signaled by
a line containing the number 0.
Output
There is exactly one line of output for each test
case. This line contains the number of days from the corresponding line of
input, followed by one blank space and the total number of gold coins paid to
the knight in the given number of days, starting with Day 1.
Sample Input
10
6
7
11
15
16
100
10000
1000
21
22
0
Sample Output
10 30
6 14
7 18
11 35
15 55
16 61
100 945
10000 942820
1000 29820
21 91
22 98
Source
Rocky Mountain 2004
简单构造法模拟，没什么说的
1 #include <iostream>
2 #include<cstdio>
3 using namespace std;
4 int a[10001];
5 int main()
6 {
7 int n,j=1,i=1,temp=1,sum=0;
8
9 for(int i = 1;i<=10000;)
10 {
11 int k = j;
12 while(k>0)
13 {
14 sum += temp;
15 a[i++] = sum;
16 k--;
17 }
18 temp++;
19 j++;
20 }
21 while(scanf("%d",&n)!=EOF&&n!=0)
22 {
23 printf("%d %d\n",n,a[n]);
24 }
25 return 0;
26 }