题目链接:http://codeforces.com/contest/34/problem/A
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — nspace-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
5 10 12 13 15 10
5 1
4 10 20 30 40
1 2
题目大意:n个士兵围成一个圆,找出相邻的两个士兵使得价值差最小
题目思路:暴力就行
代码:
#include<cstdio> #include<cmath> #include<cstring> #include<string> #include<cstdlib> #include<algorithm> #include<iostream> #include<queue> #include<stack> #include<map> using namespace std; #define FOU(i,x,y) for(int i=x;i<=y;i++) #define FOD(i,x,y) for(int i=x;i>=y;i--) #define MEM(a,val) memset(a,val,sizeof(a)) #define PI acos(-1.0) const double EXP = 1e-9; typedef long long ll; typedef unsigned long long ull; const int INF = 0x3f3f3f3f; const ll MINF = 0x3f3f3f3f3f3f3f3f; const double DINF = 0xffffffffffff; const int mod = 1e9+7; const int N = 1e6+5; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); std::ios::sync_with_stdio(false); int n; int a[105]; cin>>n; for(int i=1;i<=n;i++) cin>>a[i]; int minn=INF; int loc; for(int i=1;i<n;i++) { if(abs(a[i]-a[i+1])<minn) { minn=abs(a[i]-a[i+1]); loc=i; } } if(abs(a[n]-a[1])<minn) { cout<<n<<" 1"<<endl; } else cout<<loc<<" "<<loc+1<<endl; return 0; }