『壹』 求幾C語言個小游戲代碼,簡單的,要注釋、、謝謝了、
// Calcu24.cpp : Defines the entry point for the console application.
//
/*
6-6
24點游戲
*/
#include "conio.h"
#include "stdlib.h"
#include "time.h"
#include "math.h"
#include "string.h"/*
從一副撲克牌中,任取4張。
2-10 按其點數計算(為了表示方便10用T表示),J,Q,K,A 統一按 1 計算
要求通過加減乘除四則運算得到數字 24。
本程序可以隨機抽取紙牌,並用試探法求解。
*/void GivePuzzle(char* buf)
{
char card[] = {'A','2','3','4','5','6','7','8','9','T','J','Q','K'}; for(int i=0; i<4; i++){
buf[i] = card[rand() % 13];
}
}
void shuffle(char * buf)
{
for(int i=0; i<5; i++){
int k = rand() % 4;
char t = buf[k];
buf[k] = buf[0];
buf[0] = t;
}
}
int GetCardValue(int c)
{
if(c=='T') return 10;
if(c>='0' && c<='9') return c - '0';
return 1;
}
char GetOper(int n)
{
switch(n)
{
case 0:
return '+';
case 1:
return '-';
case 2:
return '*';
case 3:
return '/';
} return ' ';
}double MyCalcu(double op1, double op2, int oper)
{
switch(oper)
{
case 0:
return op1 + op2;
case 1:
return op1 - op2;
case 2:
return op1 * op2;
case 3:
if(fabs(op2)>0.0001)
return op1 / op2;
else
return 100000;
} return 0;
}
void MakeAnswer(char* answer, int type, char* question, int* oper)
{
char p[4][3];
for(int i=0; i<4; i++)
{
if( question[i] == 'T' )
strcpy(p[i], "10");
else
sprintf(p[i], "%c", question[i]);
}
switch(type)
{
case 0:
sprintf(answer, "%s %c (%s %c (%s %c %s))",
p[0], GetOper(oper[0]), p[1], GetOper(oper[1]), p[2], GetOper(oper[2]), p[3]);
break;
case 1:
sprintf(answer, "%s %c ((%s %c %s) %c %s)",
p[0], GetOper(oper[0]), p[1], GetOper(oper[1]), p[2], GetOper(oper[2]), p[3]);
break;
case 2:
sprintf(answer, "(%s %c %s) %c (%s %c %s)",
p[0], GetOper(oper[0]), p[1], GetOper(oper[1]), p[2], GetOper(oper[2]), p[3]);
break;
case 3:
sprintf(answer, "((%s %c %s) %c %s) %c %s",
p[0], GetOper(oper[0]), p[1], GetOper(oper[1]), p[2], GetOper(oper[2]), p[3]);
break;
case 4:
sprintf(answer, "(%s %c (%s %c %s)) %c %s",
p[0], GetOper(oper[0]), p[1], GetOper(oper[1]), p[2], GetOper(oper[2]), p[3]);
break;
}
}
bool TestResolve(char* question, int* oper, char* answer)
{
// 等待考生完成
int type[5]={0,1,2,3,4};//計算類型
double p[4];
double sum=0;
//
for(int i=0; i<4; i++) //循環取得點數
{
p[i]=GetCardValue(int(question[i]));
} for(i=0;i<5;i++)
{
MakeAnswer(answer,type[i],question,oper); //獲取可能的答案
switch(type[i])
{
case 0:
sum=MyCalcu(p[0],MyCalcu( p[1],MyCalcu(p[2], p[3], oper[2]),oper[1]),oper[0]); //A*(B*(c*D))
break;
case 1:
sum=MyCalcu(p[0],MyCalcu(MyCalcu(p[1], p[2], oper[1]),p[3],oper[2]),oper[0]); //A*((B*C)*D)
break;
case 2:
sum=MyCalcu(MyCalcu(p[0], p[1], oper[0]),MyCalcu(p[2], p[3], oper[2]),oper[1]); // (A*B)*(C*D)
break;
case 3:
sum=MyCalcu(MyCalcu(MyCalcu(p[0], p[1], oper[0]),p[2],oper[1]),p[3],oper[2]); //((A*B)*C)*D
break;
case 4:
sum=MyCalcu(MyCalcu(p[0],MyCalcu(p[1], p[2], oper[1]),oper[0]),p[3],oper[2]); //(A*(B*C))*D
break;
}
if(sum==24) return true;
}
return false;
}
/*
採用隨機試探法:就是通過隨機數字產生 加減乘除的 組合,通過大量的測試來命中的解法
提示:
1. 需要考慮用括弧控制計算次序的問題 比如:( 10 - 4 ) * ( 3 + A ), 實際上計算次序的數目是有限的:
A*(B*(c*D))
A*((B*C)*D)
(A*B)*(C*D)
((A*B)*C)*D
(A*(B*C))*D
2. 需要考慮計算結果為分數的情況:( 3 + (3 / 7) ) * 7
3. 題目中牌的位置可以任意交換
*/
bool TryResolve(char* question, char* answer)
{
int oper[3]; // 存儲運算符,0:加法 1:減法 2:乘法 3:除法
for(int i=0; i<1000 * 1000; i++)
{
// 打亂紙牌順序
shuffle(question);
// 隨機產生運算符
for(int j=0; j<3; j++)
oper[j] = rand() % 4; if( TestResolve(question, oper, answer) ) return true;
} return false;
}
int main(int argc, char* argv[])
{
// 初始化隨機種子
srand( (unsigned)time( NULL ) ); char buf1[4]; // 題目
char buf2[30]; // 解答
printf("***************************\n");
printf("計算24\n");
printf("A J Q K 均按1計算,其它按牌點計算\n");
printf("目標是:通過四則運算組合出結果:24\n");
printf("***************************\n\n");
for(;;)
{
GivePuzzle(buf1); // 出題
printf("題目:");
for(int j=0; j<4; j++){
if( buf1[j] == 'T' )
printf("10 ");
else
printf("%c ", buf1[j]);
} printf("\n按任意鍵參考答案...\n");
getch(); if( TryResolve(buf1, buf2) ) // 解題
printf("參考:%s\n", buf2);
else
printf("可能是無解...\n"); printf("按任意鍵出下一題目,x 鍵退出...\n");
if( getch() == 'x' ) break;
} return 0;
}
『貳』 C語言簡易文字冒險游戲源代碼
記憶游戲
#include<stdio.h>
#include<time.h>
#include<stdlib.h>
#include<windows.h>
#defineN10
intmain()
{inti,k,n,a[N],b[N],f=0;
srand(time(NULL));
printf("按1開始 按0退出:_");
scanf("%d",&n);
system("cls");
while(n!=0)
{for(k=0;k<N;k++)a[k]=rand()%N;
printf(" [請您牢記看到顏色的順序] ");
for(k=0;k<N;k++)
{switch(a[k])
{case0:system("color90");printf("0:淡藍色 ");break;//淡藍色
case1:system("colorf0");printf("1:白色 ");break;//白色
case2:system("colorc0");printf("2:淡紅色 ");break;//淡紅色
case3:system("colord0");printf("3:淡紫色 ");break;//淡紫色
case4:system("color80");printf("4:灰色 ");break;//灰色
case5:system("colore0");printf("5:黃色 ");break;//黃色
case6:system("color10");printf("6:藍色 ");break;//藍色
case7:system("color20");printf("7:綠色 ");break;//綠色
case8:system("color30");printf("8:淺綠色 ");break;//淺綠色
case9:system("color40");printf("9:紅色 ");break;//紅色
}
Sleep(1500);
system("colorf");//單個控制文字顏色
Sleep(100);
}
system("cls");
printf("0:淡藍色,1:白色,2:淡紅色,3:淡紫色,4:灰色,5:黃色,6:藍色7:綠色,8:淺綠色,9:紅色 ");
printf(" 請輸入顏色的順序:");
for(k=0;k<N;k++)scanf("%d",&b[k]);
for(k=0;k<N;k++)if(a[k]==b[k])f++;
if(f==0)printf("你的記憶弱爆了0 ");
elseif(f==1)printf("你的記憶有點弱1 ");
elseif(f<5)printf("你的記憶一般<5 ");
elseprintf("你的記憶力很強! ");
Sleep(2000);
system("cls");
printf(" 按0退出 按任意鍵繼續游戲: ");
scanf("%d",&n);
system("cls");
}
return0;
}
註:DEVc++運行通過,每輸入一個數字要加入一個空格。
『叄』 求一些C語言小游戲的源代碼,謝謝
「推箱子」C代碼:
#include <stdio.h>
#include <conio.h>
#include<stdlib.h>
#include<windows.h>
int m =0; //m代表第幾關
struct maps{short a[9][11]; };
struct maps map[5]={ 0,0,0,0,0,0,0,0,0,0,0, //共5關,每關9行11列
0,1,1,1,1,1,1,1,0,0,0,
0,1,0,0,0,0,0,1,1,1,0,
1,1,4,1,1,1,0,0,0,1,0, //0空地,1牆
1,5,0,0,4,0,0,4,0,1,0, //4是箱子,5是人
1,0,3,3,1,0,4,0,1,1,0, //3是目的地
1,1,3,3,1,0,0,0,1,0,0, //7是箱子在目的地(4+3)
0,1,1,1,1,1,1,1,1,0,0, //8是人在目的地(5+3)
0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,
0,0,1,1,1,1,0,0,0,0,0,
0,0,1,5,0,1,1,1,0,0,0,
0,0,1,0,4,0,0,1,0,0,0,
0,1,1,1,0,1,0,1,1,0,0,
0,1,3,1,0,1,0,0,1,0,0,
0,1,3,4,0,0,1,0,1,0,0,
0,1,3,0,0,0,4,0,1,0,0,
0,1,1,1,1,1,1,1,1,0,0,
0,0,0,0,0,0,0,0,0,0,0,
0,0,0,1,1,1,1,1,1,1,0,
0,0,1,1,0,0,1,0,5,1,0,
0,0,1,0,0,0,1,0,0,1,0,
0,0,1,4,0,4,0,4,0,1,0,
0,0,1,0,4,1,1,0,0,1,0,
1,1,1,0,4,0,1,0,1,1,0,
1,3,3,3,3,3,0,0,1,0,0,
1,1,1,1,1,1,1,1,1,0,0,
0,1,1,1,1,1,1,1,1,1,0,
0,1,0,0,1,1,0,0,0,1,0,
0,1,0,0,0,4,0,0,0,1,0,
0,1,4,0,1,1,1,0,4,1,0,
0,1,0,1,3,3,3,1,0,1,0,
1,1,0,1,3,3,3,1,0,1,1,
1,0,4,0,0,4,0,0,4,0,1,
1,0,0,0,0,0,1,0,5,0,1,
1,1,1,1,1,1,1,1,1,1,1,
0,0,0,0,0,0,0,0,0,0,0,
0,0,0,1,1,1,1,1,1,0,0,
0,1,1,1,0,0,0,0,1,0,0,
1,1,3,0,4,1,1,0,1,1,0,
1,3,3,4,0,4,0,0,5,1,0,
1,3,3,0,4,0,4,0,1,1,0,
1,1,1,1,1,1,0,0,1,0,0,
0,0,0,0,0,1,1,1,1,0,0,
0,0,0,0,0,0,0,0,0,0,0 };
void DrMap( ) //繪制地圖
{ CONSOLE_CURSOR_INFO cursor_info={1,0}; //隱藏游標的設置
SetConsoleCursorInfo(GetStdHandle(STD_OUTPUT_HANDLE),&cursor_info);
printf(" 推箱子");
printf(" ");
for (int i = 0; i < 9; i++)
{for (int j = 0; j < 11; j++)
{switch (map[m].a[i][j])
{case 0: printf(" "); break;
case 1: printf("■"); break;
case 3: printf("◎");break;
case 4: printf("□"); break;
case 5: printf("♀"); break; //5是人
case 7: printf("□"); break; //4 + 3箱子在目的地中
case 8: printf("♀");break; // 5 + 3人在目的地中
}
}
printf(" ");
}
}
void gtxy(int x, int y) //控制游標位置的函數
{ COORD coord;
coord.X = x;
coord.Y = y;
SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE), coord);
}
void start( ) //開始游戲
{ int r, c; //人的下標
for (int i = 0; i < 9; i++)
{ for (int j = 0; j < 11; j++)
{if (map[m].a[i][j] == 5||map[m].a[i][j]==8) { r = i; c = j; } } //i j 人的下標
}
char key;
key = getch( );
switch (key)
{case 'W':
case 'w':
case 72:
if (map[m]. a[r - 1][c] == 0|| map[m]. a [r - 1][c] == 3)
{ gtxy(2*c+8,r-1+3); printf("♀"); // gtxy(2*c+8,r-1+3)是到指定位置輸出字元
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r - 1][c] += 5; map[m]. a [r][c] -= 5; }
else if (map[m]. a [r - 1][c] == 4 || map[m]. a [r - 1][c] == 7)
{ if (map[m]. a [r - 2][c] == 0 || map[m]. a [r - 2][c] == 3)
{ gtxy(2*c+8,r-2+3); printf("□"); gtxy(2*c+8,r-1+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r - 2][c] += 4; map[m]. a [r - 1][c] += 1;
map[m]. a [r][c] -= 5; }
} break;
case 'S':
case 's':
case 80:
if (map[m]. a [r + 1][c] == 0 || map[m]. a [r + 1][c] == 3)
{ gtxy(2*c+8,r+1+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r + 1][c] += 5; map[m]. a [r][c] -= 5; }
else if (map[m]. a [r + 1][c] == 4 || map[m]. a [r+ 1][c] == 7)
{ if (map[m]. a [r + 2][c] == 0 || map[m]. a [r + 2][c] == 3)
{ gtxy(2*c+8,r+2+3); printf("□"); gtxy(2*c+8,r+1+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r + 2][c] += 4; map[m]. a [r + 1][c] += 1;
map[m]. a [r][c] -= 5; }
}break;
case 'A':
case 'a':
case 75:
if (map[m]. a [r ][c - 1] == 0 || map[m]. a [r ][c - 1] == 3)
{ gtxy(2*(c-1)+8,r+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r ][c - 1] += 5; map[m]. a [r][c] -= 5; }
else if (map[m]. a [r][c - 1] == 4 || map[m]. a [r][c - 1] == 7)
{if (map[m]. a [r ][c - 2] == 0 || map[m]. a [r ][c - 2] == 3)
{ gtxy(2*(c-2)+8,r+3); printf("□"); gtxy(2*(c-1)+8,r+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r ][c - 2] += 4; map[m]. a [r ][c - 1] += 1;
map[m]. a [r][c] -= 5; }
}break;
case 'D':
case 'd':
case 77:
if (map[m]. a [r][c + 1] == 0 || map[m]. a [r][c + 1] == 3)
{ gtxy(2*(c+1)+8,r+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8) {gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r][c + 1] += 5; map[m]. a [r][c] -= 5; }
else if (map[m]. a [r][c + 1] == 4 || map[m]. a [r][c + 1] == 7)
{ if (map[m]. a [r][c + 2] == 0 || map[m]. a [r][c + 2] == 3)
{ gtxy(2*(c+2)+8,r+3); printf("□"); gtxy(2*(c+1)+8,r+3); printf("♀");
if(map[m]. a[r ][c] == 5){gtxy(2*c+8,r+3); printf(" "); }
if(map[m]. a[r ][c] == 8){gtxy(2*c+8,r+3); printf("◎");}
map[m]. a [r][c + 2] += 4; map[m]. a [r][c + 1] += 1;
map[m]. a [r][c] -= 5; }
}break;
}
}
int ifwan( ) //是否完成(1是0否)
{ if(m==0){if(map[m].a[5][2]==7&& map[m].a[5][3]==7&&
map[m].a[6][2]==7&& map[m].a[6][3]==7) return 1;}
if(m==1){if(map[m].a[5][2]==7&& map[m].a[6][2]==7&&
map[m].a[7][2]==7) return 1;}
if(m==2){if(map[m].a[7][1]==7&& map[m].a[7][2]==7&& map[m].a[7][3]==7&&
map[m].a[7][4]==7&& map[m].a[7][5]==7) return 1;}
if(m==3){if(map[m].a[4][4]==7&& map[m].a[4][5]==7&& map[m].a[4][6]==7&&
map[m].a[5][4]==7&& map[m].a[5][5]==7&& map[m].a[5][6]==7) return 1;}
if(m==4){if(map[m].a[3][2]==7&& map[m].a[4][1]==7&& map[m].a[4][2]==7&&
map[m].a[5][1]==7&& map[m].a[5][2]==7) return 1;}
return 0;
}
int main( ) //主函數
{ while (1)
{ system("cls");
DrMap( );
while (1)
{ start( );
if(ifwan()){printf("