C语言课程设计3

1、C语言课程设计3Problem B: 深入浅出学算法016-猴子吃桃Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 655 Solved: 402Description猴子第一天摘下N个桃子，当时就吃了一半，还不过瘾，就又吃了一个。第二天又将剩下的桃子吃掉一半，又多吃了一个。以后每天都吃前一天剩下的一半零一个。到第D天在想吃的时候就剩一个桃子了,求第一天共摘下来多少个桃子？Input多组测试数据，每组输入1个整数DOutput对于输入的整数D求出第一天的桃子数NSample Input2Sample Output4HINT代码：#include<

2、stdio.h>intmain()intn,t,i,s;while(scanf("%d",&n)!=EOF)t=1;s=0;for(i=0;i<n-1;i+)s=(t+1)*2;t=s;printf("%dn",t);Problem K: Climb the Taishan Mountain(II)Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 870 Solved: 546DescriptionLabor day is coming.Since the holiday is long,Man

3、y students are planning to have a tour. The Taishan Mountain is very beautiful, so WangPeng and his classmates want to climb the Taishan Mountain. Mountaineering sports is WangPeng's avocation, so he is very excited. Suddenly, a very amusing idea comes to him. The Taishan Mountain has many stair

4、s, and he can climb the mountain one step by 2 stairs or by 1 stairs. How many ways does he have to climb the mountain? For example, the stairs is 2, he has 2 ways.The first one is the first step by 1 stair, and the next step by 1 stair.And the second way is only one step by 2 stairs.InputThe input

5、is including many test cases. Every line is one test case with a integer number N(N<=40). N is the stairs of the Taishan Mountain.OutputYou should output how many ways WangPeng can climb the mountain, each case output a line.Sample Input123Sample Output123HINT#include<stdio.h> int main() in

6、t n,i,j,a41; while(scanf("%d",&n)!=EOF) a0=1; a1=2; for(i=2;i<40;i+) ai=ai-1+ai-2; printf("%dn",an-1); Problem J: 深入浅出学算法024-奶牛问题Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 418 Solved: 215Description好好过个年吧，遇雪灾了， 好好火炬传递，遇藏独了 ， 发展农村医疗吧，发手足口病了 买点股票吧，大小非减持了 做火车吧，还出轨了 在家待着吧，还地

7、震了 天将降大运于斯国也，必先下其大雪，撞其火车，抢其火炬，震其国土，涨其物价，跌其股市，空房其民。 四川大地震，牵动了中华民族的很多颗心。小明决定用实际行动来支援四川，所以他决定养奶牛给前线运牛奶。 最初他只有一头母牛，它每年年初生一头小母牛。每头小母牛从第四个年头开始，每年年初也生一头小母牛。请编程实现在第n年的时候，共有多少头母牛？Input输入数据由多个测试实例组成，每个测试实例占一行，包括一个整数n(0 < n < 55)，n的含义如题目中描述。 n=0表示输入数据的结束，不做处理。Output对于每个测试实例，输出在第n年的时候母牛的数量。 每个输出占一行。Sample

8、 Input2450Sample Output246HINT#include<stdio.h>intmain()intf57=1,1,2,3,n,i;for(i=4;i<57;i+)fi=fi-1+fi-3;while(scanf("%d",&n)!=EOF&&n!=0)printf("%dn",fn);Problem L: Lease Common MultipleTime Limit: 1 Sec Memory Limit: 64 MBSubmit: 868 Solved: 597DescriptionGiv

9、en 2 positive integer x(1<=x<=1000) and y(1<=y<=1000), you are to count the Lease Common Multiple of x and y.Inputx and y, one line for each test.Outputthe Lease Common Multiple of x and y.Sample Input12 183 9Sample Output369#include<stdio.h>intlcm(intx,inty);intgcd(intx,inty);intm

10、ain()intsum,x,y;while(scanf("%d%d",&x,&y)!=EOF)sum=lcm(x,y);printf("%dn",sum);return0;intgcd(intx,inty)if(x%y=0)returny;returngcd(y,x%y);intlcm(intx,inty)returnx*y/gcd(x,y);Problem E: 深入浅出学算法019-求n的阶乘Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 735 Solved: 473Description求

11、阶乘，采用递归的方法，你会写吗？Input多组测试数据，首先输入整数T表示组数然后每一组在一行输入一个整数n( 1 <= n <= 10)Output对于每组数据输出一行，值为n的阶乘Sample Input12Sample Output2HINT使用递归函数求n!int fact(int n)#include <stdio.h>intfact(intnum)if(1=num)return1;return(num * fact(num-1);intmain()intnum,result,T,i;scanf("%d",&T);for(i=1;i

12、<=T;i+)scanf("%d",&num);result=fact(num);printf("%dn", result);return0;Problem D: 深入浅出学算法018-兔子繁殖Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 573 Solved: 302Description兔子繁殖问题。设有一对新生的兔子，从第三个月开始他们每个月月初都生一对兔子，新生的兔子从第三个月月初开始又每个月生一对兔子。按此规律，并假定兔子没有死亡，n(n<=20)个月月末共有多少个兔子？Input

13、多组测试数据，每组输入整数nOutput每组输出一行，值为n个月后的兔子对数Sample Input3Sample Output2HINT#include <stdio.h>intf(intn);intmain(void)intx;while(scanf("%d",&x)!=EOF)printf("%dn", f(x);return0;intf(intn)if(n=1|n=2)return1;elsereturnf(n-1)+f(n-2);Problem F: 深入浅出学算法020-排队购票Time Limit: 1 Sec Memo

14、ry Limit: 64 MBSubmit: 411 Solved: 271Description一常球赛开始前，售票工作正在进行中。每张球票为50元，现有k(1 <= k <= 30)人排队购票,其中n人手持50元的钱，剩余m人手持100元的钱，假设开始时售票处没有钱，求出售票处不出现找不开钱的不同排队种数（拿同样钞票面值的人换位置视为同一种排队）Input多组测试数据，先输入整数T表示组数，然后每组输入2个整数n和mOutput对于每组测试数据输出1行，值为题目描述中的排队种数Sample Input21 21 1Sample Output01HINT#include<s

15、tdio.h>intmain(void)intm,n,i,j,T,k;scanf("%d",&T);longf100100;for(k=1;k<=T;k+)scanf("%d",&m);scanf("%d",&n);for(j=1;j<=m;j+)fj0=1;for(j=0;j<=m;j+)for(i=j+1;i<=n;i+)for(j=i;j<=m;j+)fji=fj-1i+fji-1;printf("%ldn",fmn);return0;Problem

16、 G: 深入浅出学算法021-汉诺塔问题Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 453 Solved: 264Description汉诺塔（又称河内塔）问题是源于印度一个古老传说的益智玩具。大梵天创造世界的时候做了三根金刚石柱子，在一根柱子上从下往上按照大小顺序摞着64片黄金圆盘。大梵天命令婆罗门把圆盘从下面开始按大小顺序重新摆放在另一根柱子上。并且规定，在小圆盘上不能放大圆盘，在三根柱子之间一次只能移动一个圆盘。Input输入圆盘数nOutput输出搬盘子的次数Sample Input12Sample Output13#include<

17、stdio.h>intmain(void)doublemove(intm);intn;while(scanf("%d",&n)!=EOF)if(n<=40)printf("%.0fn",move(n);elseprintf("%.4en",move(n);return0;doublemove(intm)doubles;if(m=1)s=1;elses=2*move(m-1)+1;returns;Problem H: 深入浅出学算法022-汉诺塔问题IITime Limit: 1 Sec Memory Limit:

18、64 MBSubmit: 526 Solved: 205Description汉诺塔（又称河内塔）问题是源于印度一个古老传说的益智玩具。大梵天创造世界的时候做了三根金刚石柱子，在一根柱子上从下往上按照大小顺序摞着64片黄金圆盘。大梵天命令婆罗门把圆盘从下面开始按大小顺序重新摆放在另一根柱子上。并且规定，在小圆盘上不能放大圆盘，在三根柱子之间一次只能移动一个圆盘。Input输入圆盘数n( 1 <= n <= 10)Output按照示例输出搬盘子的过程，每次搬动输出一行Sample Input2Sample Outputa->ba->cb->cHINT三根柱子分别标为a b c#include<stdio.h>voidhn(intm,chara,charb,charc);voidmv(charx,chary);intmain(void)intn;while(scanf("%d",&n)!=EOF)hn(n,'a','b','c