Codeforces Round #280 (Div. 2)

This problem is simply astonishing.

ABC is almost always a sign-in question.

D words

Convert two people's firing time into an integer

Ask for a gcd, except for a moment.

Let's say the firing frequency of the two is 1 seconds x,1 second Y.

X, y gcd is G

Conversion is equivalent to

The first person y/g second shot, the second person x/g second shot a

Then two people will shoot at the same time in x/g*y/g seconds.

So every x/g*y/g second is a cycle.

Assuming the monster's blood has a, then a% (X+y) is the last cycle to shoot the amount of blood

At this time I have been too lazy to continue to think, go straight to the number of people shooting, OK

#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cmath > #include <algorithm> #include <map> #define MAXN 55555#define maxm 222222#define INF 1000000001using    Namespace Std;int n, x, Y, a;int main () {scanf ("%d%d%d", &n, &x, &y);    int g = __GCD (x, y);    x/= g;    Y/= G;        for (int i = 0; i < n; i++) {scanf ("%d", &a);        A%= (x + y); if (a = = 0 | |        (A + 1)% (x + y) = = 0) puts ("Both");            else {int flag = 0;            int low = 1, high = x;                while (low <= high) {int mid = (low + high) >> 1;                Long Long TMP = (long Long) mid * (long long) y;                Long Long z = tmp/(long long) x;                if (z + mid > a) {high = mid-1;                    } else if (z + mid = = a) {flag = 1;                Break      } else {low = mid + 1;          }} if (flag) {puts ("Vanya");            } else {puts ("Vova"); }}} return 0;}

E words

The topic gives a very good limit.

is from the x direction or the y direction, you start from 0, walk a n step, must be able to traverse all the 0~n-1

And then X-direction you go from 0 simulation to an n-step, get an x-coordinate sequence

Y coordinates are doing the same thing.

Two sequences. can be recycled.

Then you choose to go from one point (x0,y0)

Is nothing more than the two sequence, the x sequence starts from the x0, the y sequence starts from the y0, each walks a n step, to see what the point is meets the question request

Once again, it was found that the relative position of the two sequences was seen.

For all given points, the relative positions of the corresponding two sequences are calculated, and the last statistic can be

#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cmath > #include <algorithm> #include <map> #define MAXN 55555#define maxm 222222#define INF 1000000001using    Namespace Std;int posx[1111111], Posy[1111111];int N, M, DX, Dy;int x[111111], y[111111];int num[1111111];int main () {    scanf ("%d%d%d%d", &n, &m, &AMP;DX, &dy);    int now = 0;    int ind = 0;        while (posx[now] = = 0) {Posx[now] = ++ind;    Now = (now + dx)% n;    } now = 0, ind = 0;        while (posy[now] = = 0) {Posy[now] = ++ind;    Now = (now + dy)% n;    } int mx = 0, p = 0;        for (int i = 0; i < m; i++) {scanf ("%d%d", &x[i], &y[i]);        int px = posx[x[i]];        int py = posy[y[i]];        int t = (py-px + N)% n;        num[t]++;            if (Num[t] > mx) {mx = num[t];        p = t;        }} for (int i = 0; i < m; i++) {int px = posx[x[i]]; int py =Posy[y[i]];        int t = (py-px + N)% n;            if (t = = p) {printf ("%d%d\n", X[i], y[i]);        Break }} return 0;}

Codeforces Round #280 (Div. 2)