C語(yǔ)言實(shí)例：搬山游戲

設(shè)有n座山，計(jì)算機(jī)與人為比賽的雙方，輪流搬山。規(guī)定每次搬山的數(shù)止不能超 過k座，誰(shuí)搬最后一座誰(shuí)輸。游戲開始時(shí)。計(jì)算機(jī)請(qǐng)人輸入山的總數(shù)(n)和每次允許搬山的最大數(shù)止(k)。然后請(qǐng)人開始，等人輸入了需要搬走的山的數(shù)目后，計(jì)算機(jī)馬上打印出它搬多少座山，并提示尚余多少座山。雙方輪流搬山直到最后一座山搬完為止。計(jì)算機(jī)會(huì)顯示誰(shuí)是贏家，并問人是否要繼續(xù)比賽。若人不想玩了，計(jì)算機(jī)便會(huì)統(tǒng)計(jì)出共玩了幾局，雙方勝負(fù)如何。

*問題分析與算法設(shè)計(jì)

計(jì)算機(jī)參加游戲時(shí)應(yīng)遵循下列原則：

1) 當(dāng)：

剩余山數(shù)目-1<=可移動(dòng)的最大數(shù)k 時(shí)計(jì)算機(jī)要移(剩余山數(shù)目-1)座，以便將最后一座山留給人。

2)對(duì)于任意正整數(shù)x,y，一定有：

0<=x%(y+1)<=y

在有n座山的情況下，計(jì)算機(jī)為了將最后一座山留給人，而且又要控制每次搬山的數(shù)目不超過最大數(shù)k，它應(yīng)搬山的數(shù)目要滿足下列關(guān)系：

(n-1)%(k+1)

如果算出結(jié)果為0，即整除無余數(shù)，則規(guī)定只搬1座山，以防止冒進(jìn)后發(fā)生問題。

按照這樣的規(guī)律，可編寫出游戲程序如下：

#include

int main()

{

int n,k,x,y,cc,pc,g;

printf("More Mountain Game\n");

printf("Game Begin\n");

pc=cc=0;

g=1;

for(;;)

{

printf("No.%2d game \n",g++);

printf("---------------------------------------\n");

printf("How many mpuntains are there?");

scanf("%d",&n);

if(!n) break;

printf("How many mountains are allowed to each time?");

do{

scanf("%d",&k);

if(k>n||k<1) printf("Repeat again!\n");

}while(k>n||k<1);

do{

printf("How many mountains do you wish movw away?");

scanf("%d",&x);

if(x<1||x>k||x>n) /*判斷搬山數(shù)是否符合要求*/

{

printf("IIIegal,again please!\n");

continue;

}

n-=x;

printf("There are %d mountains left now.\n",n);

if(!n)

{

printf("...............I win. You are failure...............\n\n");cc++;

}

else

{

y=(n-1)%(k+1); /*求出最佳搬山數(shù)*/

if(!y) y=1;

n-=y;

printf("Copmputer move %d mountains away.\n",y);

if(n) printf(" There are %d mountains left now.\n",n);

else

{

printf("...............I am failure. You win..................\n\n");

pc++;

}

}

}while(n);

}

printf("Games in total have been played %d.\n",cc+pc);

printf("You score is win %d,lose %d.\n",pc,cc);

printf("My score is win %d,lose %d.\n",cc,pc);

}

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