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“NotFound”输入：298199输出：1输入：298167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>例7-3源程序int main(void)( inti, flag,x;Enter 5 integers: 2 981 9int a[5];Enter x: 9printf("Enter5integers:")for(i= 0; i< 5; i++)Index is 1scanf("%d", &a[i]);printf("Enter x:");Enter 5 integers: 2 9 8 1 9scanf("%d", &x);Enter x: 7flag = 0;Not Foundfor(i= 0; i< 5; i++)if(a[1] == x)(printf("Index is %din", i);flag = 1;flag的作用？break;1if(flag == 0)printf("Not Foundin");return 0;1
#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)例 7-3 思考(1)( int i, flag, x; int a[5];printf("Enter 5 integers:");去掉break语句，结果？for(i= 0; i< 5; i++)scanf("%d", &a[i]);printf("Enter x: ");Enter5integers:29819scanf("%d", &x);Enter x: 9flag = 0;Index is 1for(i= 0; i< 5; i++)Index is 4if(a[i] == x)[printf("Index is %din", i);flag = 1;-break,7if(flag == 0)printf("Not Foundin");return O;1
#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)例 7-3 思考(2)( inti, sub, x;int a[5];printf("Enter 5 integers:");for(i= 0; i< 5; i++)Enter 5 integers: 2 9 8 1 9scanf("%d", &a[i]);Enter x: 9printf("Enter x: ");Index is 4scanf("%d", &x);sub = -1;for(i= 0; i< 5; i++)sub的作用？if(a[1] == x)sub =i;if(sub != -1) printf("Index is %din", sub);else printf("Not Foundin");return O;1
#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？
例7-4求最小值#include<stdio.h>int main(void)Enter n: 6 inti,min, n;Enter6integers:29-1816int a[10];min is -1printf("Enter n:");scanf("%d", &n);printf("Enter%d integers:", n);for(i= 0; i<n; i++)scanf("%d", &a[i]);方法！min = a[0];for(i=1;i<n; i++)if(a[]< min) min = a[i];虽得到了最小值，但不能printf("min is %d In", min);确定最小值所在下标！return O;1
#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！