标签：pos title def bsp been ring [1] dice das
After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn‘t been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i, j, k) (i < j < k), such that ai·aj·akis minimum possible, are there in the array? Help him with it!
The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains npositive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.
Print one number — the quantity of triples (i, j, k) such that i, j and k are pairwise distinct and ai·aj·ak is minimum possible.
4
1 1 1 1
4
5
1 3 2 3 4
2
6
1 3 3 1 3 2
1
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there‘s only one way to choose indices.
思路：
分三种情况来逐一考虑，
a[1]=a[2]=a[3]
a[1],a[2]=a[3]
a[1],a[2],a[3]
根据题目要求的约数，只可能为这三种情况，
分类处理下就OK
细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘\0‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ map<ll,ll> m; ll n; ll a[maxn]; ll getpc(ll m,ll n)//m 中 选 n 个 { long long ans=1; for(long long k=1; k<=n; k++) { ans=(ans*(m-n+k))/k; } return ans; } int main() { gbtb; cin>>n; repd(i,1,n) { cin>>a[i]; m[a[i]]=m[a[i]]+1; } sort(a+1,a+1+n); ll ans=0ll; set<ll> s; repd(i,1,3) { s.insert(a[i]); } // 1 1 1 1 1 if(s.size()==1) { ll sum=m[a[1]]; // c 4 3 ans=getpc(sum,3); }else if(s.size()==2) { // 1 1 2 2 2 2 3 // 1 2 2 2 2 2 ll f=m[a[1]]; if(f==1) { ans=getpc(m[a[2]],2); }else { ans=m[a[3]]; } }else { // 1 2 3 3 3 3 ans=m[a[3]]; } // cout< cout<<ans<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘\n‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
Makes And The Product CodeForces - 817B （思维+构造）
标签：pos title def bsp been ring [1] dice das
原文地址：https://www.cnblogs.com/qieqiemin/p/10328020.html
