hdu5914 Triangle 【贪心】
Triangle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1413 Accepted Submission(s): 840
Problem Description
Mr. Frog has n sticks, whose lengths are 1,2, 3⋯n respectively. Wallice is a bad man, so he does not want Mr. Frog to form a triangle with three of the sticks here. He decides to steal some sticks! Output the minimal number of sticks he should steal so that Mr. Frog cannot form a triangle with
any three of the remaining sticks.
any three of the remaining sticks.
Input
The first line contains only one integer T (T≤20), which indicates the number of test cases.
For each test case, there is only one line describing the given integer n (1≤n≤20).
For each test case, there is only one line describing the given integer n (1≤n≤20).
Output
For each test case, output one line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of sticks Wallice should steal.
Sample Input
3
4
5
6
Sample Output
Case #1: 1
Case #2: 1
Case #3: 2
题意:读入n,有n个长度为1,2,3...n的木棍,问最少拿走多少根木棍,使剩下的任意两根木棍无法构成三角形。
思路:只有剩余的数字是斐波那契数1,2,3,5,8...时,拿走的木棍数才最少,所以我们最需要用木棍总数减去此范围内斐波那契数的总数就是需要拿走的最少木棍数。这道题数据范围不大,所以我是直接循环判断,数据范围比较大的话可以考虑打表。
#include<stdio.h>
int main()
{
int first,fi,next,t,n,ans,t2=0;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
ans = 2;
first = 1;
next = 2;
while(first+next <= n)
{
fi = next + first;
ans ++;
first = next;
next = fi;
}
printf("Case #%d: ",++t2);
if(n==1||n==2||n==3)
printf("0\n");
else
printf("%d\n",n-ans);
}
return 0;
}