7-17 crawling worms (15 points)

A worm is long 1 " , In a deep for N The bottom of the well . Known worms every 1 Minutes to climb up U " , But must rest 1 Minutes to climb up . In the process of rest , The worm has slipped again D " . That's it , Go up and down again . Excuse me, , How long does it take for the worm to climb out of the well ？

There is not enough demand here 1 Minutes press 1 Minute meter , And assume that as long as the head of the worm reaches the top of the well during a climb , So the worm is done . At the beginning , Worms lie at the bottom of the well （ That is, the height is 0）.

Input format ：

The input is given in order on one line 3 A positive integer N、U、D, among D<U,N No more than 100.

Output format ：

Output the time when the worm climbed out of the well in one line , In minutes .

sample input ：

12 3 1

sample output ：

11

Their thinking : I think the difficulty of the problem is when to climb up and fall down , In fact, you will find that climbing up is time t All are 1,3,5 Odd number , And the number falling down is even , Then just judge t Whether it is a prime number or not , But we should also consider the situation that the initial upward climb is greater than or equal to the well depth , for example ：2,3,1;

The code is as follows ：

#include<stdio.h>
int main()
{
    int N,U,D,h=0,t;
    scanf("%d %d %d\n",&N,&U,&D);
    for(t=1;h<N;t++){
        if(t%2!=0){
            h+=U;
            if(h>=N){
                break;
            }
        }
        else{
            h-=D;
        }
    }
    printf("%d",t);
    return 0;
}