11.3

11.3

函式递回

#include<stdio.h>
void countto10(int)
int main(){
    countto3(1);
    return 0;
}
                  //函式递回输出一到三 
void countto3(int i){
    if(i <= 3) {
        printf("%d\n",i);
        countto(i+1);
    }
}

#include<stdio.h>
void countto1(int);
int main(){
    countto1(3);
    return 0;                //函式递回由三印到一
}
void countto1(int i){
    if(i<=3&&i>0){
        printf("%d\n",i);
        countto1(i-1);
    }
}

#include<stdio.h>
void countto1(int);
int main(void){
    countto1(1)
        return 0;
}
                 //因为函式递回，印出的仍然是三到一
void countto1(int i){
    if (i<=3){
        countto1(i+1)
            printf("%d\n",i)
    }
}

#include<stdio.h>
int sum(int);
int main(){
    int n;
    printf("输入一个数：");
    scanf("%d",&n);
    printf("%d",sum(n));
    return 0;
                            //用递回函式计算连续数字的和
} 
int sum(int a){
    if (a==1){
        return 1;
    }
    return sum(a-1)+a;
}

#include<stdio.h>
int s(int);
int main(){
    int n;
    printf("上的楼层数为：");
    scanf("%d",&n);
    printf("%d",s(n));
    return 0;
}          //用函式递回计算上楼梯方法
int s(int a){
    if(a<=2){
     return a;
    }
     return s(a-1)+s(a-2);
}

阵列变数

一群具有相同资料元素集合的资料形态

会用一段连续的记忆体来存储

元素资料形态 阵列变数名称[元素个数]

int var[3] 这个变数叫var，var是个阵列

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main(){
    srand(time(0));
    int c1=0,c2=0,c3=0,c4=0,c5=0,c6=0;
    int a;
    for(a =1;a<=6000;a++){
        int num =rand()%6+1;
    switch(num){
        case 1:c1++;break;
        case 2:c2++;break;
        case 3:c3++;break;
        case 4:c4++;break;
        case 5:c5++;break;
        case 6:c6++;break;
    }         //计算掷6000次骰子个点被掷到的次数 
  }
     printf("%d\n",c1);
      printf("%d\n",c2);
       printf("%d\n",c3);
        printf("%d\n",c4);
         printf("%d\n",c5);
          printf("%d\n",c6);
          return 0;
}

==元素资料形态 阵列变数名称[元素个数]={第一个数的值，第二个数的值……,最后一个数的值}

int num[6]={0} 六个数会都被初始化为0

int num[]={0,1,2,3,4,5} 会被默认为六个数

int num[3]={3,5} 点个书会被默认初始化为0

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main(){
    srand(time(0));
    int c[6]=0;
    int a;
    for(a =1;a<=6000;a++){
        int num =rand()%6+1;
    switch(num){
        case 1:c[0]++;break;
        case 2:c[1]++;break;
        case 3:c[2]++;break;
        case 4:c[3]++;break;
        case 5:c[4]++;break;
        case 6:c[5]++;break;
    }         //计算掷6000次骰子个点被掷到的次数 
  }
     printf("%d\n",c[0]);
      printf("%d\n",c[1]);
       printf("%d\n",c[2]);
        printf("%d\n",c[3]);
         printf("%d\n",c[4]);
          printf("%d\n",c[5]);
          return 0;
}