1 条题解
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0
C :
#include<stdio.h> #include<math.h> int main() { int n,a[99999],b[99999]; int x=1000,m=1000,c; int i,j=0; scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) { if(x>a[i]) x=a[i]; } for(i=0;i<n;i++) { if(x==a[i]) { b[j]=i+1; j++; } } for(i=1;i<n;i++) { if(b[i]!=0) { c=b[i]-b[i-1]; if(m>fabs(c)) { m=fabs(c); } } } printf("%d ",m); return 0; }C++ :
#include <bits/stdc++.h> #define int long long using namespace std; const int N=1e5+10; typedef pair<int,int> PII; typedef priority_queue<int , vector<int>, greater<int>> minqueue; typedef priority_queue<int, vector<int>, less<int>> maxqueue; int gcd(int a, int b){ return b ? gcd(b, a % b) : a; } int lcm(int a, int b){ return a * b / __gcd(a, b); } int qmi(int base, int power, int p) { int result = 1; while (power > 0) { if (power & 1) result = result * base % p; base = base * base % p ; power >>= 1; } return result % p; } int a[N]; int b[N]; void solve() { int n; cin>>n; for(int i=1;i<=n;i++) { cin>>a[i]; } int w = 1; int minn = 1e9 + 10; for(int i=1;i<=n;i++) { if(a[i] < minn) { minn = a[i]; w = 1; b[w] = i; w++; } else if(a[i] == minn) { b[w] = i; w++; } } int minx = 1e5 + 10; for(int i=2;i<w;i++) { minx = min(b[i]-b[i-1],minx); if(minx == 1) { cout<<1; return ; } } cout<<minx; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; t=1; //cin>>t; while(t--) { solve(); } return 0; }
- 1
信息
- ID
- 737
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 8
- 标签
- 提交数
- 29
- 已通过
- 5
- 上传者