有兩名球員的骰子游戲

Question

我必須與兩名球員做一個遊戲。每個投擲一個骰子，然後骰子數量更大的那個需要一個點。這發生了十次。

這個程序好嗎？我該如何做這個遊戲？

int main(int argc, char** argv) 

    int i; 
    int sumplayer1=0,sumplayer2=0; 
    int dice1 = 0; 
    int dice2 = 0; 
    time_t t; 
    srand(time(&t)); 

    for (i=0;i<=10; i++) 
    { 

    dice1 = (rand() % 6); 
    dice2 = (rand() % 6); 
     if (dice1>dice2) 
      sumplayer1=sumplayer1+1; 
     if (dice1<dice2) 
      sumplayer2=sumplayer2+1; 
     if (dice1==dice2){ 
      sumplayer1=sumplayer1; 
      sumplayer2=sumplayer2; 


    } 


      if (sumplayer1>sumplayer2){ 
       printf("player 1 won"); 
      } 
      if (sumplayer1<sumplayer2){ 
       printf("player 2 won"); 
      } 



    return 0; 

} 
} 

Answer 1

No.Your code is not okay。你有一些錯誤在這裏。這裏是糾正的代碼，所有的錯誤更正：

#include <stdio.h> 
#include <stdlib.h> 
#include <time.h> //neccessary headers 

int main(int argc, char** argv) 
{ // You forgot this 
    int i; 
    int sumplayer1=0,sumplayer2=0; 
    int dice1 = 0; 
    int dice2 = 0; 
    time_t t; 
    srand(time(&t)); 

    for (i=0;i<10; i++) // use < not <= as you want to loop 10 times 
    { 

    dice1 = (rand() % 6)+1; 
    dice2 = (rand() % 6)+1; //+1 because die values are 1-6 not 0-5 
     if (dice1>dice2) 
      sumplayer1++; 
     else if (dice1<dice2) //use else if 
      sumplayer2++; //++ is much easier to understand 
     /*if (dice1==dice2){ // No need of this part 
      sumplayer1=sumplayer1; 
      sumplayer2=sumplayer2;}*/ 
    } 
    if (sumplayer1>sumplayer2){ 
      printf("player 1 won"); 
     } 
     else if (sumplayer1<sumplayer2){ //use else if 
      printf("player 2 won"); 
     }else //There is a possibilty when both players get equal score 
      printf("Its a tie"); 


    return 0; 

} 
//} Extra one here