题目链接 https://www.lydsy.com/JudgeOnline/problem.php?id=2038
参考博客 https://www.cnblogs.com/Paul-Guderian/p/6933799.html
https://www.cnblogs.com/hzf-sbit/p/4056874.html
解析 C(2,n)=n*(n-1)/2=(n*n-n)/2
分子就是∑C(sum[i],2) ( i 是区间内的颜色 sum[i] 是区间内i颜色的数量）公式真难写。。。 还是看第二篇博客吧。
1 /************************************************************** 2 Problem: 2038 3 User: 1071532391 4 Language: C++ 5 Result: Accepted 6 Time:1708 ms 7 Memory:5604 kb 8 ****************************************************************/ 9 10 #include <bits/stdc++.h> 11 #define pb push_back 12 #define mp make_pair 13 #define fi first 14 #define se second 15 #define all(a) (a).begin(), (a).end() 16 #define fillchar(a, x) memset(a, x, sizeof(a)) 17 #define huan printf(" "); 18 #define debug(a,b) cout<<a<<" "<<b<<" "; 19 using namespace std; 20 typedef long long ll; 21 const int maxn=1e5+10,inf=0x3f3f3f3f; 22 const ll mod=1e9+7; 23 ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;} 24 int n,m,unit,col[maxn],be[maxn]; 25 ll sum[maxn],ans=0; 26 struct node 27 { 28 int l,r,id; 29 ll ans1,ans2; 30 }a[maxn]; 31 int cmp1(node a,node b) 32 { 33 if(be[a.l]==be[b.l]) 34 return a.r<b.r; 35 return a.l<b.l; 36 } 37 int cmp2(node a,node b) 38 { 39 return a.id<b.id; 40 } 41 ll S(ll x) {return x*x;} 42 void revise(int x,int add) 43 { 44 ans-=S(sum[col[x]]); 45 sum[col[x]]+=add; 46 ans+=S(sum[col[x]]); 47 } 48 int main() 49 { 50 scanf("%d%d",&n,&m); 51 unit=sqrt(n); 52 for(int i=1;i<=n;i++) 53 { 54 scanf("%d",col+i);be[i]=i/unit+1; 55 } 56 for(int i=1;i<=m;i++) 57 { 58 scanf("%d%d",&a[i].l,&a[i].r); 59 a[i].id=i; 60 } 61 sort(a+1,a+m+1,cmp1); 62 int l=1,r=0; 63 for(int i=1;i<=m;i++) 64 { 65 while(l<a[i].l)revise(l,-1),l++; 66 while(l>a[i].l)revise(l-1,1),l--; 67 while(r>a[i].r)revise(r,-1),r--; 68 while(r<a[i].r)revise(r+1,1),r++; 69 if(a[i].l==a[i].r) 70 { 71 a[i].ans1=0,a[i].ans2=1; 72 continue; 73 } 74 a[i].ans1=ans-(a[i].r-a[i].l+1); 75 a[i].ans2=1ll*(a[i].r-a[i].l+1)*(a[i].r-a[i].l); 76 ll gcd_=gcd(a[i].ans1,a[i].ans2); 77 a[i].ans1/=gcd_;a[i].ans2/=gcd_; 78 } 79 sort(a+1,a+m+1,cmp2); 80 for(int i=1;i<=m;i++) 81 printf("%lld/%lld ",a[i].ans1,a[i].ans2); 82 return 0; 83 } 84