1 条题解
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0
C :
#include<stdio.h> int main() { int n,m[1000]; int i, j, k; int max, min; scanf("%d", &n); for ( i = 0; i < n; i++) { scanf("%d", &m[i]); } max = min = m[0]; for (j = 0; j < n; j++) { if (m[j] > max) max = m[j]; } for (k = 0; k < n; k++) { if (m[k] < min) min = m[k]; } printf("%d %d\n", max, min); }
C++ :
#include <iostream> #include <cstdio> using namespace std; int main() { int n,num,max_num,min_num; cin>>n; cin>>num; max_num=num; min_num=num; for(int i=2;i<=n;i++) { cin>>num; if(max_num<num) { max_num=num; } if(min_num>num) { min_num=num; } } cout<<max_num<<" "<<min_num; return 0; }
Pascal :
var a:array[1..1000]of integer; n,i,z,x:integer; begin readln(n); for i:=1 to n do read(a[i]); z:=a[1];x:=a[1]; for i:=1 to n do begin if z<a[i] then z:=a[i]; if x>a[i] then x:=a[i]; end; writeln(z,' ',x); end.
- 1
信息
- ID
- 427
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者