2016多校联合第5场部分题解 HDU5780,5781,5783,5784,5785,5787,5791,5792

5780

∑gcd(xa−1,xb−1)这个式子可以把里面拆开成多项式，变式为∑(x−1)∗gcd(1+x+...+xa−1,1+x+...+xb−1)

然后由于gcd(1+x+...+xa−1,1+x+...+xb−1)=(1+x+...+xgcd(a,b))

然后再带入化简求和在化简得到∑gcd(xa−1,xb−1)=∑xgcd(a,b)−1

对于求gcd=d的对数，可以考虑欧拉函数，然后对数就是2∗(phi[1]+...+phi[nd])−1。预处理出来这些东西，直接枚举d会超时。考虑n/d有很多共同的变量，那么我们只需要枚举这sqrt(n)个不同的变量，然后x求个等比公式就可以在sqrt(n)的时间内搞定了

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const ll MOD = 1e9 + 7;const int maxn = 1e6 + 50;const int maxv = 1e3 + 10;const double eps = 1e-9;ll phi[maxn];void exgcd(ll a, ll b, ll &d, ll &x, ll &y) {    if(!b) {d = a; x = 1; y = 0;}    else {exgcd(b, a%b, d, y, x); y -= x * (a / b);}}ll inv(ll a, ll n) {    ll d, x, y;    exgcd(a, n, d, x, y);    return d == 1 ? (x + n) % n : -1;}void init() {    phi[1] = 1;    phi[0] = 0;    for(int i = 2; i <= 1e6 + 5; i++) {        if(!phi[i]) {            for(int j = i; j <= 1e6 + 5; j += i) {                if(!phi[j]) phi[j] = j;                phi[j] = phi[j] / i * (i - 1);            }        }    }    for(int i = 1; i <= 1e6 + 5; i++) {        phi[i] += phi[i-1];    }    for(int i = 1; i <= 1e6; i++) {        phi[i] = (2 * phi[i] - 1 + MOD) % MOD;    }}ll Pow(ll a, ll n) {    ll ret = 1;    while(n) {        if(n & 1) ret = ret * a % MOD;        a = a * a % MOD;        n >>= 1;    }    return ret;}ll calc(ll a, ll b, ll x) {    ll ret = 0;    ret = (Pow(x, b+1) - Pow(x, a) + MOD) % MOD;    ret = ret * inv(x - 1, MOD) % MOD;    // printf("ret = %lld\n", ret);    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    init();    int T;    scanf("%d", &T);    while(T--) {        ll x, n;        scanf("%lld%lld", &x, &n);        ll res = MOD - n * n % MOD;        if(x == 1) {puts("0"); continue;}        for(int i = 1, la = 0; i <= n; i = la + 1) {            la = n / (n / i);            // printf("a = %d  b = %d\n", i, la);            res = (res + calc(i, la, x) * phi[n/i] % MOD) % MOD;        }        printf("%lld\n", res);    }    return 0;}

HDU5781

题目原型是一个选楼层扔鸡蛋找到敲好不碎的楼层。这里取钱和扔鸡蛋一样，被警告一次等同于碎一个蛋。设f(i,j)表示当前可能的钱为(0,i)还剩j次被警告的机会找到准确楼层的最小期望。

对于当前状态，显然可以考虑枚举当前我们取的钱，从1开始（0无意义）到i，然后根据题目条件，钱数是均匀分布的，那么对于当前决策有两种可能，被警告或者不被警告。若被警告，说明钱数小于k，这样的概率是ki+1，若不被警告，说明钱数大于等于k，概率为i+1−ki+1

这样dp[i][j]=min(ki+1∗dp[k−1][j−1]，i+1−ki+1∗dp[i−k][j])+1

直接计算是n2m但是想想因为最优策略，至少是接近2分的策略，那么钦定m≤15即可

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const double inf = 1000000000000.0;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 2e3 + 10;const double eps = 1e-9;double dp[maxv][20];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int k, w;    k = 2000; w = 15;    w = min(w, 15);    for(int i = 0; i <= k; i++)    for(int j = 0; j <= w; j++)    dp[i][j] = inf;    CLR(dp[0]);    for(int i = 1; i <= k; i++) {        for(int j = 1; j <= w; j++) {            for(int k = 1; k <= i; k++) {                dp[i][j] = min(dp[i][j], (i-k+1.0)/(i+1.0)*dp[i-k][j] + k/(i+1.0)*dp[k-1][j-1] + 1.0);            }        }    }    while(scanf("%d%d", &k, &w) != EOF) {        w = min(w, 15);        printf("%.6f\n", dp[k][w]);    }       return 0;}

HDU5783

因为要求分割区间的各前缀和大于等于0，那么考虑倒着来，先遇到的0和正数显然独立成为一个区间，当遇到某些连续的负数就去要用前面的正数和0去填补了。

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;ll a[maxn];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int n;    while(scanf("%d", &n) != EOF) {        for(int i = 1; i <= n; i++) {            scanf("%lld", a + i);        }        int ans = 0;        ll sum = 0;        for(int i = n; i > 0; i--) {            sum += a[i];            if(sum >= 0) {                ans++;                sum = 0;            }        }        printf("%d\n", ans);    }    return 0;}

HDU5784

要求点集中锐角三角形的个数，一个锐角三角形3锐角，一个直角和钝角都是2锐角，搞出所有锐角个数，然后减去直角和钝角的两倍就是锐角三角形的锐角个数，然后除以3就是锐角三角形个数。

为了减小时间开销，可以枚举一个点，然后以它为中心，其他点做极角排序，先扫出锐角的个数，在扫出小于PI的角个数，一减就是钝角+直角个数。因为要往后扩PI，可以把原数组复制一遍，注意精度，eps至少10−12

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 2e3 + 10;const double eps = 1e-12;const double PI = acos(-1);ll x[maxv], y[maxv];int comp(double x) {    if(abs(x) < eps) return 0;    return x < 0 ? -1 : 1;}int m;ll calc(std::vector<double> &a, double delta) {    ll ret = 0;    for(int i = 0, j = 0, k = 0; i < m; i = k + 1) {        while(k + 1 < m && comp(a[k+1] - a[i]) == 0) ++k;        j = max(j, k);        while(j < a.size() && comp(a[j] - a[i] - delta) < 0) ++j;        ret += (k - i + 1) * (j - k - 1);    }    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int n;    // cout << atan2(1, 2e9) * 180 << endl;    while(scanf("%d", &n) != EOF) {        for(int i = 1; i <= n; i++) {            scanf("%lld%lld", &x[i], &y[i]);        }        ll rui = 0, dun = 0;        for(int T = 1; T <= n; T++) {            std::vector<double> a;            for(int j = 1; j <= n; j++) {                if(T == j) continue;                a.push_back(atan2(y[j]-y[T], x[j]-x[T]));            }            m = a.size();            for(int i = 0; i < m; i++) {                a.push_back(PI * 2 + a[i]);            }            sort(ALL(a));            int tot = 0, curAcute = 0;            curAcute = calc(a, PI / 2);            tot = calc(a, PI);            rui += curAcute;            dun += tot - curAcute;        }        printf("%lld\n", (rui - dun * 2) / 3);    }    return 0;}

HDU5785

此题介绍两种做法。

一种是题解说的，用manacher处理出各个最长的回文串的区间，就是下标算的蛋疼。在考虑一个区间对答案的影响的时候，设这个区间为[l,mid,r],mid是区间的中心，可以发现在mid到l,r两端的时候，l+r−i的值都是当前坐标i对应的回文串的端点。在[l,mid]就是右端点，[mid,r]就是左端点，在处理右端点数组的时候，就在l位置加上l+r在mid+1位置减去l+r。并且考虑当前点被区间覆盖的个数num也在相应的地方加1和减1，然后求个前缀和便能算出当前点右端点贡献。详细细节见代码

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 2e6 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;char a[maxn], s[maxn*2];int n, p[maxn*2];void init() {    s[0] = '@'; s[1] = '#', n = 2;    for(int i = 0; a[i]; i++) {        s[n++] = a[i];        s[n++] = '#';    }    s[n] = 0;}void kp() {    init();    int mx = 0, id;    for(int i = 1; i < n; i++) {        if(mx > i) p[i] = min(mx - i, p[id * 2 - i]);        else p[i] = 1;        while(s[i - p[i]] == s[i + p[i]]) p[i]++;        if(i + p[i] > mx) {            mx = i + p[i];            id = i;        }    }}ll L[maxn], R[maxn];ll lsum[maxn], rsum[maxn];ll lnum[maxn], rnum[maxn];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    while(scanf("%s", a) != EOF) {        kp(); n--;        CLR(lnum); CLR(rnum);        CLR(L); CLR(R); CLR(lsum); CLR(rsum);        // for(int i = 1; i <= n; i++) {        //  printf("%c", s[i]);        // }        // puts("");        for(int i = 2; i < n; i++) {            if(i & 1) {                int cur = i / 2;                int l = cur + 1 - (p[i] - 1) / 2;                int r = cur + (p[i] - 1) / 2;                if(r <= l) continue;                int back = cur + 1;                // printf("%d %d %d %d\n", l, cur, back, r);                lsum[back] += l + r, lnum[back]++;                 lsum[r+1] -= l + r, lnum[r+1]--;                rsum[l] += l + r, rnum[l]++;                rsum[cur+1] -= l + r, rnum[cur+1]--;            }            else {                int cur = i / 2;                int l = cur - (p[i] - 2) / 2;                int r = cur + (p[i] - 2) / 2;                // printf("%d %d %d\n", l, cur, r);                lsum[cur] += l + r, lnum[cur]++;                lsum[r+1] -= l + r, lnum[r+1]--;                rsum[l] += l + r, rnum[l]++;                rsum[cur+1] -= l + r, rnum[cur+1]--;            }        }        ll res = 0;        n /= 2;        for(int i = 1; i <= n; i++) {            lsum[i] += lsum[i-1];            rsum[i] += rsum[i-1];            lnum[i] += lnum[i-1];            rnum[i] += rnum[i-1];        }        // for(int i = 1; i <= n; i++) {        //  printf("lsum[%d] = %lld\n", i, lsum[i] - i);        //  printf("rsum[%d] = %lld\n", i, rsum[i] - i);        // }        for(int i = 1; i <= n; i++) {            L[i] = (lsum[i] - i * lnum[i] % MOD + MOD) % MOD;            R[i] = (rsum[i] - i * rnum[i] % MOD + MOD) % MOD;        }        for(int i = 1; i < n; i++) {            res = (res + L[i] * R[i+1] % MOD) % MOD;        }        printf("%lld\n", res);    }    return 0;}

另一种是用回文树处理，方便了很多，但是注意空间限制。我们设当前处理的点是i−1与i，那么对答案的贡献就是∑a∑b(i−a)(i+b−1)其中a,b分别是以i−1结尾的各个回文串长度，以i开头的各个回文串长度。

设cnta,suma,cntb,sumb分别是以i结尾的回文串个数，长度和，以i开头的回文串个数，长度和。

拆开后为

cnta∗cntb∗i2+i∗(sumb−cntb)∗cnta−i∗cntb∗suma−suma∗sumb

回文树可以统计出以i下标结尾的回文串个数，回文串长度和，顺着倒着分别搞一次，cnta,cntb,suma,sumb都出来了带入计算即可。

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 18 + 10;const double eps = 1e-9;const int SIZE = 26;struct PalindromicTree {    //nxt指针，指向的串为当前串加上同一个字符构成    int nxt[maxn][SIZE];    //fail指针，失配后跳转到fail指向的节点，指向的节点为当前节点的最长公共回文后缀    int fail[maxn];    int cnt[maxn];    int num[maxn];    int len[maxn];    char S[maxn];    int last;    int n;    int p;    int newNode(int l) {        for(int i = 0; i < SIZE; i++) nxt[p][i] = 0;        cnt[p] = num[p] = 0;        len[p] = l;        return p++;    }    void init() {        p = 0;        newNode(0);        newNode(-1);        last = 0;        n = 0;        S[n] = -1;        fail[0] = 1;    }    int getFail(int x) {        while(S[n - len[x] - 1] != S[n]) x = fail[x];        return x;    }    pair <ll, ll> add(int c) {        c -= 'a';        S[++n] = c;        int cur = getFail(last);        if(!nxt[cur][c]) {            int now = newNode(len[cur] + 2);            fail[now] = nxt[getFail(fail[cur])][c];            nxt[cur][c] = now;            num[now] = num[fail[now]] + len[now];            if(num[now] >= MOD) num[now] -= MOD;            cnt[now] = cnt[fail[now]] + 1;        }        last = nxt[cur][c];        return {cnt[last], num[last]};    }}pt;char s[maxn];int Lcnt[maxn], Lnum[maxn];ll mul(ll a, ll b) {    return a * b % MOD;}int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    // freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    while(scanf("%s", s + 1) != EOF) {        int n = strlen(s+1);        pt.init();        for(int i = 1; i <= n; i++) {            auto ret = pt.add(s[i]);            Lcnt[i] = ret.first;            Lnum[i] = ret.second;        }        pt.init();        ll ans = 0;        for(int i = n; i > 1; i--) {            auto ret = pt.add(s[i]);            ll sufcnt = ret.first;            ll sufnum = ret.second;            ll L = mul(mul(i, i), mul(sufcnt, Lcnt[i-1]));            ll M = mul(Lcnt[i-1], sufnum - sufcnt) - mul(sufcnt, Lnum[i-1]) + MOD;            M %= MOD;            M = mul(M, i);            ll R = mul(sufnum - sufcnt, Lnum[i-1]);            ans += L + M - R;            ans = (ans + MOD) % MOD;        }        printf("%lld\n", ans);    }    return 0;}

HDU5787

明显的数位dp。题解很机智，直接map套vector方便了很多。因为K最大才是5，显然保存前4位，往下走的时候要看当前位能不能填数字i。这样边界问题比较蛋疼，前导零处理起来比较麻烦。于是我又开了一维表示当前需要的与之前比较的位数。最后为了只memset一次又开了一维表示当前处理的K是多少。。具体细节见代码

#include <set>#include <map>#include <queue>#include <stack>#include <ctime>#include <cmath>#include <string>#include <cctype>#include <cstdio>#include <vector>#include <bitset>#include <cstdlib>#include <cstring>#include <iomanip>#include <sstream>#include <iostream>#include <assert.h>#include <algorithm>using namespace std;#define ls id<<1,l,mid#define rs id<<1|1,mid+1,r#define OFF(x) memset(x,-1,sizeof x)#define CLR(x) memset(x,0,sizeof x)#define MEM(x) memset(x,0x3f,sizeof x)typedef long long ll;typedef pair<int,int> pii;const int maxn = 1e4 + 50;const int maxm = 1e6 + 50;const double eps = 1e-10;const int max_index = 62;const int inf = 0x3f3f3f3f ;int MOD = 1e5;ll dp[19][maxn][6][4];int digit[20];ll L, R, K;int all;ll dfs(int pos, int num, int bit, int limit) {    if(pos == -1) return 1;    if(!limit && dp[pos][num][bit][K-2] != -1) return dp[pos][num][bit][K-2];    int up = limit ? digit[pos] : 9;    ll ret = 0;    int tmp = num;    int viss[10];    CLR(viss);    for(int i = 0; i < bit; i++) {        viss[tmp % 10]++;        tmp /= 10;    }    for(int i = 0; i <= up; i++) {        if(viss[i]) {            continue;        }        int nbit = bit;        if(bit == 0 && i == 0) nbit = 0;        else nbit++;        if(nbit >= K) nbit--;        ret += dfs(pos-1, (num*10+i)%MOD, nbit, i == up && limit);    }    if(!limit) dp[pos][num][bit][K-2] = ret;    return ret;}ll calc(ll x) {    int pos = 0;    while(x) {        digit[pos++] = x % 10;        x /= 10;    }    all = pos - 1;    return dfs(pos-1, 0, 0, 1);}int main () {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif    // cout << sizeof(dp) / 1024 << endl;    memset(dp, -1, sizeof dp);    while(scanf("%lld%lld%lld", &L, &R, &K) != EOF) {        MOD = 1;        for(int i = 1; i < K; i++) MOD *= 10;        printf("%lld\n", calc(R) - calc(L - 1));    }    return 0;}

HDU5791

dp水题，dp[i][j]表示s1串前i个字符，s2串前j个字符匹配的序列的总个数。

显然有dp[i][j]=dp[i−1][j]+dp[i][j−1]−dp[i−1][j−1]因为此处dp[i−1][j−1]被dp[i−1][j],dp[i][j−1]包含，多算了一次所以需要减去。然后若s1[i]==s2[j]说明可以以i,j为底构建新的序列，这时候dp[i][j]+=dp[i−1][j−1]+1

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;ll dp[maxv][maxv];int a[maxv], b[maxv];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int n, m;    while(scanf("%d%d", &n, &m) != EOF) {        for(int i = 1; i <= n; i++) {            scanf("%d", a + i);        }        for(int i = 1; i <= m; i++) {            scanf("%d", b + i);        }        CLR(dp);        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= m; j++) {                dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1];                dp[i][j] = (dp[i][j] % MOD + MOD) % MOD;                // printf("dp[%d][%d] = %lld\n", i, j, dp[i][j]);                if(a[i] == b[j]) {                    dp[i][j] = (dp[i][j] + dp[i-1][j-1] + 1) % MOD;                }                // printf("dp[%d][%d] = %lld\n", i, j, dp[i][j]);            }        }        printf("%lld\n", dp[n][m]);    }    return 0;}

HDU5792

要找出满足要求的四元组，可以先处理出所有顺序对个数，逆序对个数，然后去重。设Aa,Ab,Ac,Ad其中a<b,c<d,Aa<Ab,Ac<Ad重复就是当Aa==Ac or Aa==Ad or Ab==Ac or Ab==Ad四种情况。统计可以用bit

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e5 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;int lowbit(int x) {return x & (-x);}int MAXN;void updata(ll *a, int id, int val) {    while(id <= MAXN) {        a[id] += val;        id += lowbit(id);    }}ll query(ll *a, int id) {    ll ret = 0;    while(id) {        ret += a[id];        id -= lowbit(id);    }    return ret;}ll A[maxn];ll dp[maxn];ll dpv[maxn];ll lbig[maxn], rbig[maxn], lsmall[maxn], rsmall[maxn];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int n;    while(scanf("%d", &n) != EOF) {        std::vector<int> xs;        for(int i = 1; i <= n; i++) {            scanf("%d", A + i);            xs.push_back(A[i]);        }        sort(ALL(xs));        xs.resize(unique(ALL(xs)) - xs.begin());        MAXN = xs.size() + 5;        for(int i = 0; i <= MAXN; i++) {            dp[i] = dpv[i] = 0;        }        int SIZE = xs.size() + 2;        ll small = 0;        ll big = 0;        CLR(lbig); CLR(rbig); CLR(lsmall); CLR(rsmall);        for(int i = n; i > 0; i--) {            int id = lower_bound(ALL(xs), A[i]) - xs.begin() + 1;            ll L = query(dp, id - 1);            ll R = query(dpv, SIZE - id - 1);            updata(dp, id, 1);            updata(dpv, SIZE - id, 1);            rbig[i] = R;            rsmall[i] = L;            small += L;            big += R;        }        // printf("small = %lld big = %lld\n", small, big);        for(int i = 0; i <= MAXN; i++) {            dp[i] = dpv[i] = 0;        }        for(int i = 1; i <= n; i++) {            int id = lower_bound(ALL(xs), A[i]) - xs.begin() + 1;            ll L = query(dp, id - 1);            ll R = query(dpv, SIZE - id - 1);            updata(dp, id, 1);            updata(dpv, SIZE - id, 1);            lbig[i] = R;            lsmall[i] = L;        }        ll ans = small * big;        for(int i = 1; i <= n; i++) {            ans = ans-rbig[i]*rsmall[i]-lbig[i]*lsmall[i]-lsmall[i]*rsmall[i]-lbig[i]*rbig[i];        }        printf("%lld\n", ans);    }    return 0;}