Codeforces Round #282 (Div. 2)

Also hundreds of years did not do CF, this game is still done after the game, a lot of degradation

A.digital Counter

There may be a match stick lost, ask you how many of the number of possible numbers, only the naked eye to see each number of the match stick may form a number, a table on the line

#include <iostream>#include<cstdio>using namespacestd;Const intPos[] = {2,7,2,3,3,4,2,5,1,2};intA, B;intMain () {#ifdef LOCAL freopen ("495a.in","R", stdin);#endifa= GetChar ()-'0'; b= GetChar ()-'0'; printf ("%d\n", pos[a]*Pos[b]); return 0;}

B. Modular equations

At first did not read test instructions, thought is the same yu ... It is

a/x = y ... b gives you a A, a, a, you can ask for the number of X

Because B is the remainder, it needs to meet 0<=b<x

And the vegetarian number only needs to be checked to sqrt (x) almost, here only to enumerate to sqrt (A-B)

This complexity is still acceptable O (sqrt (A-B))

At first it was silly to enumerate from sqrt (A-B) to A-B ... Instant explosion ... TLE

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>using namespacestd;Const intMAXN =1e5;intA, B, CNT;intPCNT[MAXN];BOOLPRIME[MAXN];intMain () {#ifdef LOCAL freopen ("495b.in","R", stdin);#endifscanf ("%d%d", &a, &b); if(A < b) printf ("0\n"); Else if(A = = b) printf ("infinity\n"); Else{a-=b;  for(inti =1, t = sqrt (a); I <= t; i++) {            if(A% i)Continue; intx = i, y = A/i; if(x > B) cnt++; if(Y > B) cnt++; if(x > b && x = = y) cnt--; } printf ("%d\n", CNT); }    return 0;}

C. Treasure

The feeling is greedy ... At the beginning of the idea of a silly, but also to subdivide a lot of discussion, and put themselves in the

is to let the front ' # ' fill in 1 ') ', and the last one to fill in the remaining required ') '

Run and check if there's a place that's going to make ' the number redundant ' (' number

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue  >using namespace Std;const int maxn = 1e5+10;int mcnt, acnt, Hpos, Lpos, lh;intinf[maxn];char ins[maxn];bool possible = True;queue<int> h;void Noanswer () {printf (" -1\n"); exit (0);} int main () {#ifdef localfreopen ("495c.in", "R", stdin), #endifscanf ("%s", ins), for (int i = 0, t = strlen (ins); i < T; i++ {if (ins[i] = = ' (') mcnt++;else if (ins[i] = = ') ') mcnt--;else {lh = i;mcnt--;inf[i] = 1;h.push (i);}} if (mcnt < 0) noanswer () INF[LH] + = MCNT;MCNT = 0;for (int i = 0, t = strlen (ins); i < T; i++) {if (ins[i] = = ' (') mcnt++ ; else if (ins[i] = = ') ') mcnt--;else mcnt-= inf[i];if (mcnt < 0) noanswer ();} while (!h.empty ()) {printf ("%d\n", Inf[h.front ()]); H.pop ();} return 0;}

　

D.obsessive String

I feel that the D problem of CF is enough to sanction me ... KMP long forgotten, not to mention the DP, you have time to write it ~

ABC to feel the details are still quite a lot of, do not know because they degenerate, or the game partial details

Codeforces Round #282 (Div. 2)