Norma
Time Limit: 20 Sec Memory Limit: 64 MB[Submit][Status][Discuss]
Description
Input
第1行，一个整数N；
第2~n+1行，每行一个整数表示序列a。
Output
输出答案对10^9取模后的结果。
Sample Input
4
2
4
1
4
2
4
1
4
Sample Output
109
HINT
N <= 500000
1 <= a_i <= 10^8
Solution
1 egin {align} 2 &基本思路，考虑分治，统计对于左端点在[L,mid]，右端点在[mid+1，R]的区间贡献：\ 3 \ 4 &枚举在[L, mid]中的点i，[mid+1,R]中的点a、b，\ 5 &保证min[i,mid]≤min[mid+1,a]\ 6 &并且max[i,mid]≥max[mid+1,b] \ 7 &然后[mid+1,R]就被分成了三部分：[mid+1,a],[a+1,b],[b+1,R]，(假设a<b)，\ 8 \ 9 &我们考虑分别对三个部分计算贡献：(max=max[i,mid], min=min[i,mid])\ 10 &①.[mid+1,a]:Ans=max*min*sum_{j=mid+1}^{a}(j-i+1) \ 11 &②.[a+1,b]:Ans=max*sum_{j=a+1}^{b}(min[a+1,j]*(j-i+1)) \ 12 &③.[b+1,R]:Ans=sum_{j=b+1}^{R}( max[b+1,j]*min[b+1,j]*(j-i+1)) \ 13 \ 14 &这时候，显然①可求了。我们继续推导(把(j-i+1)拆开)，显然有：\ 15 &②=max*sum_{j=a+1}^{b}min[a+1,j]*j-max*sum_{j=a+1}^{b}min[a+1,j]*(i-1)\ 16 &③=sum_{j=b+1}^{R}max[b+1,j]*min[b+1,j]*j-sum_{j=b+1}^{R}max[b+1,j]*min[b+1,j]*(i-1)\ 17 \ 18 &显然此时②中的sum_{j=a+1}^{b}min[a+1,j]*j、sum_{j=a+1}^{b}min[a+1,j]是可以O(len)预处理的，\ 19 &③中的sum_{j=b+1}^{R}max[b+1,j]*min[b+1,j]*j、sum_{j=b+1}^{R}max[b+1,j]*min[b+1,j]也可以O(len)预处理。\ 20 \ 21 &显然，若b<a类似。所以我们分治一下即可。\ 22 \ 23 &::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::——BearChild 24 25 end {align}
Code
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cmath> 8 using namespace std; 9 typedef long long s64; 10 11 const int ONE = 1000005; 12 const int MOD = 1e9; 13 const int INF = 2147483640; 14 15 int get() 16 { 17 int res = 1, Q = 1; char c; 18 while( (c = getchar()) < 48 || c > 57) 19 if(c == '-') Q = -1; 20 if(Q) res = c - 48; 21 while( (c = getchar()) >= 48 && c <= 57) 22 res = res * 10 + c - 48; 23 return res * Q; 24 } 25 26 27 int n; 28 int val[ONE]; 29 int Ans; 30 31 void Modit(int &a) 32 { 33 if(a < 0) a += MOD; 34 if(a >= MOD) a -= MOD; 35 } 36 37 struct power 38 { 39 int Min, MinI; 40 int Max, MaxI; 41 int MinMax, MinMaxI; 42 43 friend power operator -(power a, power b) 44 { 45 power c; 46 Modit(c.Min = a.Min - b.Min), 47 Modit(c.MinI = a.MinI - b.MinI), 48 Modit(c.Max = a.Max - b.Max), 49 Modit(c.MaxI = a.MaxI - b.MaxI), 50 Modit(c.MinMax = a.MinMax - b.MinMax), 51 Modit(c.MinMaxI = a.MinMaxI - b.MinMaxI); 52 return c; 53 } 54 }A[ONE]; 55 56 57 int Sum(int a, int b) 58 { 59 return (s64)(a + b) * (b - a + 1) / 2 % MOD; 60 } 61 62 void Solve(int L, int R) 63 { 64 if(L == R) 65 { 66 Modit(Ans += (s64)val[L] * val[R] % MOD * 1); 67 return; 68 } 69 70 int mid = L + R >> 1; 71 72 int minn = INF, maxx = -INF; 73 A[mid] = (power){0, 0, 0, 0, 0, 0}; 74 75 for(int j = mid + 1; j <= R; j++) 76 { 77 minn = min(minn, val[j]), maxx = max(maxx, val[j]), 78 Modit(A[j].Min = A[j - 1].Min + minn), 79 Modit(A[j].MinI = A[j - 1].MinI + (s64)minn * j % MOD), 80 Modit(A[j].Max = A[j - 1].Max + maxx), 81 Modit(A[j].MaxI = A[j - 1].MaxI + (s64)maxx * j % MOD), 82 Modit(A[j].MinMax = A[j - 1].MinMax + (s64)minn * maxx % MOD), 83 Modit(A[j].MinMaxI = A[j - 1].MinMaxI + (s64)minn * maxx % MOD * j % MOD); 84 } 85 86 minn = INF, maxx = -INF; 87 int a = mid, b = mid; 88 power del; 89 90 for(int i = mid; i >= L; i--) 91 { 92 minn = min(minn, val[i]), maxx = max(maxx, val[i]); 93 while(minn <= val[a + 1] && a + 1 <= R) a++; 94 while(maxx >= val[b + 1] && b + 1 <= R) b++; 95 if(a <= b) 96 { 97 Modit(Ans += (s64)maxx * minn % MOD * Sum(mid + 1 - i + 1, a - i + 1) % MOD); 98 del = A[b] - A[a]; 99 Modit(Ans += (s64)maxx * del.MinI % MOD - (s64)maxx * del.Min % MOD * (i - 1) % MOD); 100 del = A[R] - A[b]; 101 Modit(Ans += (s64)del.MinMaxI - (s64)del.MinMax * (i - 1) % MOD); 102 } 103 else 104 { 105 Modit(Ans += (s64)maxx * minn % MOD * Sum(mid + 1 - i + 1, b - i + 1) % MOD); 106 del = A[a] - A[b]; 107 Modit(Ans += (s64)minn * del.MaxI % MOD - (s64)minn * del.Max % MOD * (i - 1) % MOD); 108 del = A[R] - A[a]; 109 Modit(Ans += (s64)del.MinMaxI - (s64)del.MinMax * (i - 1) % MOD); 110 } 111 } 112 113 Solve(L, mid), Solve(mid + 1, R); 114 115 } 116 117 int main() 118 { 119 n = get(); 120 for(int i = 1; i <= n; i++) 121 val[i] = get(); 122 Solve(1, n); 123 printf("%d", Ans); 124 }