Codeforces Round #328 (Div. 2)

This CF, ready to go back to the bedroom to take a bath, sleep, can result ...

Water a-pawnchess

The first time you forget to judge the equality of a first go to a win, hack off. Later I learned that my code was wrong (fall

for (int i=1; i<=8; ++i) {scanf ("%s", S[i]);    ！！！ }

Mathematics (Law-seeking) b-the Monster and the Squirrel

Test instructions: Each vertex of a polygon is referred to another point, and if it touches another ray, it stops and asks how many areas the final polygon is divided into.

Analysis: Move the following:

problem B. The Monster and the squirrel

After drawing the rays from the first vertex (n -2) triangles is formed. The subsequent rays would generate independently sub-regions in these triangles. Let's analyse the triangle determined by vertices 1, i, i + 1, after drawing the rays from vertex i and (i + 1) The triangle would be divided into (n - i) + (i-2) = n -2 regions. Therefore the total number of convex regions is (n -2)2

If The squirrel starts from the region that has 1 as a vertex, then she can go through each region of Triangle (1 , i, i + 1) once. That implies, the squirrel can collect all the walnuts in (n -2)2 jumps.

I've been guessing the conclusion and tried some formulas, but the situation with 3,4 is not right. (Stroke

Pay attention to explode int

Math c-the Big Race

Test instructions: Two in the race, each person's speed fixed, the end <=t, ask two people dead heat possible situation

Analysis: Classification discussion, if the LCM (b, W) <= T, then the points of multiples of each LCM and then followed by Min (b, W)-1 are dead heat, other cases are not analyzed in detail ...

Note that the LCM may explode long long, which can be compared with T or converted to a double log function.

/************************************************* author:running_time* Created time:2015/10/31 Saturday 22:41:05* Fi Le name:c.cpp ************************************************/#include <cstdio> #include <algorithm># Include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 | 1typedef long ll;const int N = 1e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const Double EPS = 1e-10;c Onst Double PI = ACOs ( -1.0); ll GCD (ll A, ll b) {return B? GCD (b, a% B): A;} int main (void) {ll T, W, b;scanf ("%i64d%i64d%i64d", &t, &w, &b), if (W > B) Swap (W, b), if (W > t &amp ;& B > t) {printf ("1/1\n");} else if (B > t) {ll x = GCD (w-1, T);p rintf ("%i64d/%i64d\n", (w-1)/x, t/x);} else if (log (double) w) + log ((double) b)-Log ((double) GCD (W, b)) > Log ((Double) t)) {ll x = GCD (w-1, T);p Rin TF ("%i64d/%i64d\n", (w-1)/x, t/x);} Else{ll LCM = B/GCD (W, b) * W;IF (t% LCM = = 0) {ll y = (w-1) + (t/lcm-1) * W + 1;ll x = GCD (y, T);p rintf ("%i64d /%i64d\n ", y/x, t/x);}  Else{ll y = (w-1) + (t/lcm-1) * W + (1 + min (t LCM, w-1)), ll x = GCD (y, T);p rintf ("%i64d/%i64d\n", y/x, T/   x);}}    cout << "time Elapsed:" << 1.0 * Clock ()/clocks_per_sec << "s.\n"; return 0;}

　　

Codeforces Round #328 (Div. 2)