Topic links: https://atcoder.jp/contests/abc140/tasks/abc140_e
Subject to the effect
Given the arrangement of a 1 ~ N P.
Defined $ X_ {L, R} $ value $ P_L, P_ {L + 1}, ldots, P_R $ second largest value \ numbers.
求$\displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}$.
analysis
This is a question of obtaining the contribution, which is to be calculated for each number x as the second-largest contribution to the time of the answer.
In order to calculate the contribution of x answer, we need to know the x position Lpos1 left of the first large numbers, x left the second largest number of positions of Lpos2, the second largest of the large number of first position x and x to the right of the right Rpos1 location Rpos2 numbers.
Knowing this position after four x contribution of the answer can immediately find out.
How demand is necessary to use a static list, the detail in the code itself.
code show as below
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // ?? x ?????? c 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 24 25 #define ms0(a) memset(a,0,sizeof(a)) 26 #define msI(a) memset(a,inf,sizeof(a)) 27 #define msM(a) memset(a,-1,sizeof(a)) 28 29 #define MP make_pair 30 #define PB push_back 31 #define ft first 32 #define sd second 33 34 template<typename T1, typename T2> 35 istream &operator>>(istream &in, pair<T1, T2> &p) { 36 in >> p.first >> p.second; 37 return in; 38 } 39 40 template<typename T> 41 istream &operator>>(istream &in, vector<T> &v) { 42 for (auto &x: v) 43 in >> x; 44 return in; 45 } 46 47 template<typename T> 48 ostream &operator<<(ostream &out, vector<T> &v) { 49 Rep(i, v.size()) out << v[i] << " \n"[i == v.size()]; 50 return out; 51 } 52 53 template<typename T1, typename T2> 54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 55 out << "[" << p.first << ", " << p.second << "]" << "\n"; 56 return out; 57 } 58 59 inline int gc(){ 60 static const int BUF = 1e7; 61 static char buf[BUF], *bg = buf + BUF, *ed = bg; 62 63 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 64 return *bg++; 65 } 66 67 inline int ri(){ 68 int x = 0, f = 1, c = gc(); 69 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 70 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 71 return x*f; 72 } 73 74 template<class T> 75 inline string toString(T x) { 76 ostringstream sout; 77 sout << x; 78 return sout.str(); 79 } 80 81 inline int toInt(string s) { 82 int v; 83 istringstream sin(s); 84 sin >> v; 85 return v; 86 } 87 88 //min <= aim <= max 89 template<typename T> 90 inline bool BETWEEN(const T aim, const T min, const T max) { 91 return min <= aim && aim <= max; 92 } 93 94 typedef long long LL; 95 typedef unsigned long long uLL; 96 typedef pair< double, double > PDD; 97 typedef pair< int, int > PII; 98 typedef pair< int, PII > PIPII; 99 typedef pair< string, int > PSI; 100 typedef pair< int, PSI > PIPSI; 101 typedef set< int > SI; 102 typedef set< PII > SPII; 103 typedef vector< int > VI; 104 typedef vector< double > VD; 105 typedef vector< VI > VVI; 106 typedef vector< SI > VSI; 107 typedef vector< PII > VPII; 108 typedef map< int, int > MII; 109 typedef map< int, string > MIS; 110 typedef map< int, PII > MIPII; 111 typedef map< PII, int > MPIII; 112 typedef map< string, int > MSI; 113 typedef map< string, string > MSS; 114 typedef map< PII, string > MPIIS; 115 typedef map< PII, PII > MPIIPII; 116 typedef multimap< int, int > MMII; 117 typedef multimap< string, int > MMSI; 118 //typedef unordered_map< int, int > uMII; 119 typedef pair< LL, LL > PLL; 120 typedef vector< LL > VL; 121 typedef vector< VL > VVL; 122 typedef priority_queue< int > PQIMax; 123 typedef priority_queue< int, VI, greater< int > > PQIMin; 124 const double EPS = 1e-8; 125 const LL inf = 0x7fffffff; 126 const LL infLL = 0x7fffffffffffffffLL; 127 const LL MOD = 1E9 + . 7 ; 128 const int maxN = 1E5 + . 7 ; 129 const LL ONE = . 1 ; 130. const LL evenBits = 0xaaaaaaaaaaaaaaaa ; 131 is const LL oddBits = 0x5555555555555555 ; 132 133 int N; 134 int P [maxN]; // P array is a 1 to N arranged in groups of 135 int POS [maxN]; //POS [i] for the i position in the array P 136 int leftFirstMaxPos [maxN]; // number of positions to the left of the first position of the number i is greater than i, leftFirstMaxPos [i], if none 0 137 int rightFirstMaxPos [ maxN]; // number of positions to the right position number i of i is greater than a first rightFirstMaxPos [i], compared with no. 1 + N 138 LL ANS = 0 ; 139 140 int main () { 141 is // the freopen ( "MyOutput.txt", "W", stdout); 142 // The freopen ( "input.txt", "R & lt", stdin); 143 // the INIT (); 144 Scanf ( " % D " , & N); 145 the For (i, 1 , N) { 146 Scanf ( " % D " , & P [I]); 147 POS [P [I]] = I; 148 // followed by enumeration starts from 1, and 1 is the smallest, so that just about a recent large It is two adjacent numbers 149 leftFirstMaxPos [i] = i - . 1 ; 150 rightFirstMaxPos [i] = i + . 1 ; 151 } 152 153 // calculate the contribution of each of the second largest number of i is the time of the answer 154 the for (i, . 1 , N) { 155 int Lpos1 = leftFirstMaxPos [POS [i]]; // left i is greater than the first number of his position 156 int Lpos2 = leftFirstMaxPos [Lpos1]; // i is greater than the second position on the left his number 157 int Rpos1 = rightFirstMaxPos [POS [i]]; // the right of i is greater than the first number of his position 158 int Rpos2 = rightFirstMaxPos [Rpos1]; / / I is greater than the right side of the second position of his number 159 160. IF (! Lpos1 = 0 ) = + ANS oNE * I * (Lpos1 - Lpos2) * (Rpos1 - POS [I]); 161 IF (Rpos1 =! + N . 1 ) ANS = ONE * + I * (POS [I] - Lpos1) * (Rpos2 - Rpos1); 162 163 // update 164 / * 165 for example, this fragment: 166 . 1. 5. 4. 3 2. 6. 7 167 is 1 2. 6. 3. 5 P. 4. 7 168 In this case i = 1 to Example 169 would leftFirstMaxPos [4] = 3, rightFirstMaxPos [2] = 3 to 170 After computing the contributions i = 1, leftFirstMaxPos [4] = 2, rightFirstMaxPos [2]. 4 = 171 is that equivalent to the step 1 to update the cut, the rest of the number is the minimum number of 2, then about 2 and just recently a large number of adjacent two 172 right, operating much like a linked list of deleted values 173 leftFirstMaxPos precursor equivalent static array of linked list 174 rightFirstMaxPos equivalent static list of successor array 175 * / 176 leftFirstMaxPos [Rpos1] = Lpos1; 177 rightFirstMaxPos [Lpos1] = Rpos1; 178 } 179 180 printf("%lld", ans); 181 return 0; 182 }