Bestcoder Round #40 A B C

A, water problem, direct enumeration to SQRT

b, each time for each enumeration, if less than the current bit, then the answer can be calculated, add the answer is: Set 3 parts, the former exactly the same part A, the middle of a new selection of B, followed by the whole arrangement of C, so that each part and each two parts can be composed of the number of reverse order, because n only 100, Inside the enumeration is also no problem, mainly in the back of the full array of C reverse logarithm, this can be processed by DP, dp[i] = dp[i-1] * i + I! * SUM (i-1), sum (i) represents the sum of 1 to I.

C: Push formula, C (n, m) = C (n-1, m-1) + C (n-1, m) = C (n-1, m-1) + C (n-2, m-1) + C (n-2, m) .... So for C (i, K), a <= i <= B's sum, is equal to C (b + 1, K + 1)-C (A, K + 1), and then because the P is relatively small, to use Lucas to calculate the number of combinations can be

Code:

A:

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;int T, n;int Main () {    scanf ("%d", &t);    while (t--) {        scanf ("%d", &n);        int ans = 2000000000LL;        for (int i = 1; I * i <= n; i++) {            if (n% i = = 0) {                ans = min (ans, (n/i + i) * 2);            }        }        printf ("%d\n", ans);    }    return 0;}

B:

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef long ll; const int N = 105;const int MOD = 1000000007;int N, a[n], vis[n], dp[n], fac[n], sum[n];int tot;int dfs (int u) {if (U    = = N) return 0;    int ans = 0;    int cnt = 0;        for (int i = 1; i < a[u]; i++) {if (vis[i]) continue;        Ans = (ans + (LL) tot * fac[n-u-1]% mod)% MoD;        if (n-u-1 > 0) ans = (ans + (LL) CNT * (n-u-1)% mod * fac[n-u-2]% mod)% MoD;        Ans = (ans + dp[n-u-1])% MOD;        int tmp = 0;        for (int j = u-1; J >= 0; j--) {if (A[j] > i) tmp++;        } int sb = 0;        int SBB = 0;            for (int j = 1; J <= N; j + +) {if (j = = i) continue;            if (Vis[j]) SBB + = SB;        else sb++;        } if (N-u-1 > 0) ans = (ans + (LL) SBB * (n-u-1)% mod * fac[n-u-2]% mod)% MoD; Ans = (ans + (LL) tmp * fac[n-u-1]% mod)% MoD;        cnt++;    } Vis[a[u]] = 1;    for (int i = a[u] + 1; I <= n; i++) if (vis[i]) tot++;    Ans = (ans + DFS (U + 1))% MOD; return ans;}    int main () {fac[0] = 1;        for (int i = 1; i < N; i++) {Fac[i] = (LL) fac[i-1] * I% MOD;    Sum[i] = (Sum[i-1] + i)% MOD;    } for (int i = 2; i < N; i++) Dp[i] = ((LL) dp[i-1] * I% mod + (LL) fac[i-1] * sum[i-1]% mod)% MoD;        while (~SCANF ("%d", &n)) {for (int i = 0; i < n; i++) scanf ("%d", &a[i]);        memset (Vis, 0, sizeof (VIS));        tot = 0;    printf ("%d\n", DFS (0)); } return 0;}

C:

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int N = 100005;    typedef long Long Ll;int f[n], f2[n];int x1, x2, y1, y2, p;int pow_mod (int x, int k) {int ans = 1;        while (k) {if (k&1) ans = (ll) ans * x% P;        x = (ll) x * x% P;    K >>= 1; } return ans;    int C (int n, int m) {if (M > n | | n < 0 | | m < 0) return 0; Return (LL) f[n] * F2[m]% p * f2[n-m]% p;}    int Lucas (int n, int m) {if (M = = 0) return 1; Return (LL) C (n% p, m% p) * Lucas (n/p, m/p)% p;}        int main () {while (~scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &p)) {f[0] = 1;        for (int i = 1; i < min (N, p); i++) F[i] = (LL) f[i-1] * I% P;        F2[min (n, p)-1] = Pow_mod (F[min (n, p)-1], p-2);        for (int i = min (N, p)-2; I >= 0; i--) f2[i] = (LL) F2[i + 1] * (i + 1)% P;        int ans = 0; for (int i = y1; I <= y2; i++) ans = (ans + ((Lucas (x2 + 1, i + 1)-Lucas (x1, i + 1)% p + p)% p)% p;    printf ("%d\n", ans); } return 0;}

Bestcoder Round #40 A B C