12 条题解
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23
已纠正,感谢 小青呱(赵紫轩) (hetao2934713) !!!
这一道题简简单单得啦~看看吧!
- 首先用 max 和 min 分别求出最大值和最小值,并且用数组 a 存储数据,然后输出最大值减去最小值的差
- 然后再一次遍历数组 a 然后如果是最小值的全部输出循环变量 + 空格,即可……AC!!!!
好了,就是这么简单,来看看代码吧~
#include <bits/stdc++.h> using namespace std; int main() { int n, a[101], maxx = 0, minn = 9999999; cin >> n; for (int i = 1; i <= n; i++) { cin >> a[i]; if (a[i] > maxx) maxx = a[i]; if (a[i] < minn) minn = a[i]; } cout << maxx - minn << endl; for (int i = 1; i <= n; i++) if (a[i] == minn) cout << i << " "; return 0; }
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4
#include <iostream> using namespace std; int main() { int n,da=0,xiao=999; cin >> n; int a[n]; for (int i=0;i<n;i++) { cin >> a[i]; if (a[i]>da) { da=a[i]; } if (a[i]<xiao) { xiao=a[i]; } } cout << da-xiao << endl; for (int i=0;i<n;i++) { if (a[i]==xiao) { cout << i+1 << ' '; } } return 0; }
已AC
看前点个赞,养成好习惯!
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2
这是一道简单的数组问题 1.定义一个数组a来存储每天的温度,求出最大值和最小值。 2.输出最大值-最小值的差。 3.如果a[i]等于最小值,输出下标i。
#include <iostream> using namespace std; int main() { int n,a[105],maxn=0,mino=1000; cin >> n; for (int i=1;i<=n;i++) { cin >> a[i]; if (a[i]>maxn) { maxn=a[i]; } if (a[i]<mino) { mino=a[i]; } } cout << maxn-mino << endl; for (int i=1;i<=n;i++) { if (a[i]==mino) { cout << i << " "; } } return 0; }
已AC
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1
** c++(04)已经AC过了,放心食用
#include <bits/stdc++.h> using namespace std; int main() { int n, a[101], maxx = 0, minn = 9999999; cin >> n; for (int i = 1; i <= n; i++) { cin >> a[i]; if (a[i] > maxx) maxx = a[i]; if (a[i] < minn) minn = a[i]; } cout << maxx - minn << endl; for (int i = 1; i <= n; i++) if (a[i] == minn) cout << i << " "; return 0; } ```**
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1
过辣!
#include <bits/stdc++.h> using namespace std; int main() { int n,a[200],nnax=-114514,nnin=114514,pos; scanf("%d",&n); for (int i=1;i<=n;i++) { scanf("%d",&a[i]); if (a[i]>nnax)nnax=a[i]; if (a[i]<nnin)nnin=a[i]; } printf("%d\n",nnax-nnin); for (int i=1;i<=n;i++) { if (a[i]==nnin)printf("%d ",i); } return 0; }
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0
这道题这么简单随便看看吧放心全AC
#include <iostream> using namespace std; int main() { int n,a[105],maxx=0,minn=10000; cin>>n; for (int i=1;i<=n;i++) { cin>>a[i]; } for (int i=1;i<=n;i++) { maxx=max(maxx,a[i]); minn=min(minn,a[i]); } cout<<maxx-minn<<endl; for (int i=1;i<=n;i++) { if (a[i]==minn) { cout<<i<<" "; } } return 0; }
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0
#include <bits/stdc++.h> using namespace std; int main() { int n, h[101]; cin >> n; for (int i = 1; i <= n; i++) { cin >> h[i]; } int maxx = -1, minn = 1000; for (int i = 1; i <= n; i++) { maxx = max(maxx, h[i]); minn = min(minn, h[i]); } cout << maxx - minn << endl; for (int i = 1; i <= n; i++) { if (h[i] == minn) { cout << i << " "; } } return 0; }
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0
设置两个变量,maxx存放最大值,minn存放最小值,maxx初始值要尽可能小,minn初始值要尽可能大;
遍历n次,每次输入当天的温度,并判断是否比当前最大值大,或者比当前最小值小,若比最大值大,则更新maxx;若比最小值小,则更新minn;
循环结束后,输出最高温度和最低温度的差值;
再遍历for循环,如果当天温度等于最低温度,则输出这是第几天,即下标i;
for (int i = 1; i <= n; i++) //输入n天的温度 { cin >> a[i]; //输入当天温度 if (a[i] > maxx) //如果当天温度大于当前最大值,则更新maxx { maxx = a[i]; } if (a[i] < minn) //如果当天温度小于当前最小值,则更新minn { minn = a[i]; } } cout << maxx - minn << endl; //输出最高温度和最低温度的差值 for (int i = 1; i <= n; i++) //遍历这n天的温度 { if (a[i] == minn) //如果这天温度等于最低温度,输出i { cout << i << " "; } }
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-1
#include <bits/stdc++.h> using namespace std; int main() { int n, a[101], max = 0, min = 9999999; cin >> n; for (int i = 1; i <= n; i++) { cin >> a[i]; if (a[i] > max) max = a[i]; if (a[i] < min) min = a[i]; } cout << max - min << endl; for (int i = 1; i <= n; i++) if (a[i] == min) cout << i << " "; return 0; } 已AC
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-1
- 设置两个变量,max存放最大值,min存放最小值,max初始值要尽可能小,min初始值要尽可能大;
- 遍历n次,每次输入当天的温度,并判断是否比当前最大值大,或者比当前最小值小,若比最大值大,则更新max;若比最小值小,则更新min;
- 循环结束后,输出最高温度和最低温度的差值
- 再遍历for循环,如果当天温度等于最低温度,则输出这是第几天,即下标 i+1;
别忘了输出空格哦!
呼,终于讲完了,那废话少说,上代码吧!(UP AC过,请放心食用)
#include <bits/stdc++.h> using namespace std; int main() { int n, a[100], max = 0, min = 2147483647;//int类型的最大值 cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; if (a[i] > max)//判断最大值 { max = a[i]; } if (a[i] < min)//判断最小值 { min = a[i]; } } cout << max-min << endl;//输出最大值 for (int i = 0; i < n; i++) { if (a[i] == min) { cout << i+1 << ' ';//输出最小值 } } return 0; }
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信息
- ID
- 814
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- 2
- 标签
- 递交数
- 1100
- 已通过
- 684
- 上传者