题目链接：https://vjudge.net/contest/361581#problem/A
题目大意：问你一个区间内选两个数，有多大的概率选到两个同样的数
算概率的方法很简单，设区间内相同的数的数量分别为n1, n2....，那么概率就是(C(n1,2) + C(n2,2) + .....) / C(l-r+1,2)然后约分就是了
那么问题转化为求(C(n1,2) + C(n2,2) + .....)，我们观察发现C(2,2) = 1, C(3,2) = 3, C(4, 2) = 6，那么，我们大胆的推测(瞎猜)n每增加1， 其组合数
就增加n-1。之后再套莫队就能轻松做出来了！
#include<set> #include<map> #include<list> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl '\n' #define rtl rt<<1 #define rtr rt<<1|1 #define lson rt<<1, l, mid #define rson rt<<1|1, mid+1, r #define maxx(a, b) (a > b ? a : b) #define minn(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) memset(a, 0x3f, sizeof(a)) #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; typedef pair<ll, ll> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 1e5+10; int arr[maxn], pos[maxn], cnt[maxn]; ll res; struct Q { int l, r, id; ll ans1, ans2; } q[maxn]; inline void add(int p) { res += cnt[arr[p]]++; } inline void sub(int p) { res -= --cnt[arr[p]]; } bool cmp1(Q x, Q y) { return pos[x.l]==pos[y.l] ? x.r < y.r : pos[x.l] < pos[y.l]; } bool cmp2(Q x, Q y) { return x.id < y.id; } int main(void) { int n, m; scanf("%d%d", &n, &m); int siz = sqrt(n)+eps; for (int i = 1; i<=n; ++i) { scanf("%d", &arr[i]); pos[i] = i/siz; } for (int i = 0; i<m; ++i) { scanf("%d%d", &q[i].l, &q[i].r); q[i].id = i; } sort(q, q+m, cmp1); int l = 1, r = 0; for (int i = 0; i<m; ++i) { while(q[i].l < l) add(--l); while(q[i].r > r) add(++r); while(q[i].l > l) sub(l++); while(q[i].r < r) sub(r--); ll t = q[i].r - q[i].l; q[i].ans1 = res/__gcd(res, (t+1)*t/2); q[i].ans2 = t*(t+1)/2/__gcd(res, (t+1)*t/2); if (q[i].ans1 == q[i].ans2) q[i].ans1 = q[i].ans2 = 1; //如果上下两个数相等会得到0/0，那么结果就是1/1 } sort(q, q+m, cmp2); for (int i = 0; i<m; ++i) printf("%lld/%lld\n", q[i].ans1, q[i].ans2); return 0; }