例7-3在数组中查找一个给定的数 输入5个整数,将它们存入数组a中,再输入1个数x, 然后在数组中查找x,如果找到,输出相应的下标, 否则,输出“ Not found 输入:29819 输出:1 输入:29816 输出: Not found
输入5个整数，将它们存入数组a中，再输入1个数x， 然后在数组中查找x，如果找到，输出相应的下标， 否则，输出“Not Found”。 输入：2 9 8 1 9 9 输出：1 输入：2 9 8 1 6 7 输出：Not Found 例7-3 在数组中查找一个给定的数
include <stdio h> int main(void) 例7-3源程序 i int i, flag, X; int a[5] Enter 5 integers: 2981 9 printf("Enter 5 integers: ) Enter x: 9 for(i=0;i<5; 1++) Index is 1 scanf(%d",&a0) printf("Enter x: scanf("%d, &x) Enter 5 integers:2981 9 flag=0; Enter x: 7 ior(i=0;i<5;i++) Not Found if(a(==xr printf("Index is %dIn",i) flag=1; break: flag的作用? if(flag ==0 printf( Not Found\n); return 0
#include <stdio.h> int main(void) { int i, flag, x; int a[5]; printf(“Enter 5 integers: "); for(i = 0; i < 5; i++) scanf("%d", &a[i]); printf(“Enter x: "); scanf("%d", &x); flag = 0; for(i = 0; i < 5; i++) if(a[i] == x){ printf("Index is %d\n", i); flag = 1; break; } if(flag == 0) printf("Not Found\n"); return 0; } 例 7-3 源程序 Enter 5 integers: 2 9 8 1 9 Enter x: 9 Index is 1 Enter 5 integers: 2 9 8 1 9 Enter x: 7 Not Found flag的作用？
include <stdio. h: int main(void) i int i, flag, x; int a[5]: 例7-3思考(1) printf( Enter 5 integers: ) for(i=0; 1<5; 1++) 去掉 break语句,结果? scanf( %d", &a[] printf((“ Enter x:") scanf(%d,&x) Enter 5 integers: 29819 flag=0 Enter x: 9 for(=0;i<5;i++) Index is 1 if(a[==X) Index is 4 printf(" Index is %dIn", i flag=1; if(flag ==0) printf(" not Found\n") return o
#include <stdio.h> int main(void) { int i, flag, x; int a[5]; printf(“Enter 5 integers: "); for(i = 0; i < 5; i++) scanf("%d", &a[i]); printf(“Enter x: "); scanf("%d", &x); flag = 0; for(i = 0; i < 5; i++) if(a[i] == x){ printf("Index is %d\n", i); flag = 1; break; } if(flag == 0) printf("Not Found\n"); return 0; } 例 7-3 思考(1) Enter 5 integers: 2 9 8 1 9 Enter x: 9 Index is 1 Index is 4 去掉break语句，结果？
include <stdio. h> int main(void) i int i, sub, X 例7-3思考(2) int a5]; printf(Enter 5 integers: ) for(i=0;i<5;i++) scanf( %d", &a[i Enter 5 integers: 2981 9 printf((“ Enter x:") Enter x: 9 scanf( %d",&x); Index is 4 sub = -1 for(ⅰ=0;i<5;i+t) if(a[==x) sub的作用? sub=i if(sub =-1 )printf("Index is %dn", sub) else printf("not Foundn") return 0
#include <stdio.h> int main(void) { int i, sub, x; int a[5]; printf(“Enter 5 integers: "); for(i = 0; i < 5; i++) scanf("%d", &a[i]); printf(“Enter x: "); scanf("%d", &x); sub = -1; for(i = 0; i < 5; i++) if(a[i] == x) sub = i; if(sub != -1) printf("Index is %d\n", sub); else printf("Not Found\n"); return 0; } 例 7-3 思考(2) Enter 5 integers: 2 9 8 1 9 Enter x: 9 Index is 4 sub的作用？
include <stdio. h> 例7-4求最小值 int main(void) i int i, min, Enter n: 6 int a[101 Enter 6 integers: 2 9-1 6 printf((“ Enter n:" min is-1 scanf( %d",&n); printf("Enter %d integers: , n) for(ⅰ=0;i<n;i+t) scanf(%d", &a[i]) 方法 min= a[0 for(ⅰ=1;i<n;i+t) if(a]< min)min=a[ 虽得到了最小值,但不能 printf("min is%dmn",mn;确定最小值所在下标! return 0:
#include <stdio.h> int main(void) { int i, min, n; int a[10]; printf(“Enter n: "); scanf("%d", &n); printf(“Enter %d integers: ", n); for(i = 0; i < n; i++) scanf("%d", &a[i]); min = a[0]; for(i = 1; i < n; i++) if(a[i] < min) min = a[i]; printf("min is %d \n", min); return 0; } 例 7-4 求最小值 Enter n: 6 Enter 6 integers: 2 9 -1 8 1 6 min is -1 方法！！ 虽得到了最小值，但不能 确定最小值所在下标！