51nod1712 區間求和

http://www.51nod.com/Challenge/Problem.html#problemId=1712
先考慮題面中的簡化問題。
對於\(i\in [1,n]\)\(a_i\)的貢獻爲\(a_i*(i-1)-a_i*(n-i)\)
那麼對於\(i\in [l,r](a_l=a_r)\),貢獻爲\(a_i*(i-l)-a_i*(r-i)=2i*a_i-a_i(l+r)\)
這個式子只須要統計4個東西就能夠\(O(n)\)計算了。html

  • \(i\)左側\(a_l\)的個數\(A\)
  • \(i\)右側\(a_r\)的個數\(B\)
  • \(i\)左側\(a_l\)\(l\)之和\(C\)
  • \(i\)右側\(a_r\)\(r\)之和\(D\)

那麼對答案的貢獻就是\(2*i*a_i*A*B-a_i*(BC+AD)\)
統計一下\(A,B,C,D\)便可。c++

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return x * f;
}

typedef unsigned int uint;
const int N = 1000010;

int n, lim, a[N];
uint num[N], sum[N], numl[N], numr[N], suml[N], sumr[N];
uint Sum, Num;

int main() {
    n = read();
    for(int i = 1; i <= n; ++i) {
        a[i] = read();
        numr[a[i]]++; sumr[a[i]] += i; 
    }
    uint ans = 0;
    for(int i = 1; i <= n; ++i) {
        numl[a[i]]++; suml[a[i]] += i;
        Num -= num[a[i]];
        num[a[i]] = numl[a[i]] * numr[a[i]];
        Num += num[a[i]];
        Sum -= sum[a[i]];
        sum[a[i]] = suml[a[i]] * numr[a[i]] + sumr[a[i]] * numl[a[i]];
        Sum += sum[a[i]];
        
        ans += 2 * i * a[i] * Num - a[i] * Sum;
        
        numr[a[i]]--; sumr[a[i]] -= i;
        Num -= num[a[i]];
        num[a[i]] = numl[a[i]] * numr[a[i]];
        Num += num[a[i]];
        Sum -= sum[a[i]];
        sum[a[i]] = suml[a[i]] * numr[a[i]] + sumr[a[i]] * numl[a[i]];
        Sum += sum[a[i]];
    }
    printf("%u\n", ans);
}
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