Bestcoder Round #75 Problem Solving report

1.King ' s Cake

#include <stdio.h>#include<iostream>using namespacestd;intans;voidSolveintLintW) {    if(Weight = =0|| Weight = =0)        return 0; intT,TL,TW; T= l/W; Ans+=T; TW=W; W= l%W; L=tw; Solve (l,w);}intMainvoid){    intLength,weight,cas; CIN>>CAs;  while(cas--)    {        intT; Ans=0; CIN>> length >>Weigth; T=length; if(Length <weight) {Length=weight; Weight=T;        } solve (Length,weigth); cout>> ans >>Endl; }    return 0;}

2.King ' s Phone

#include <iostream>#include<string.h>#include<stdio.h>using namespacestd;intm[Ten],a[3][3];voidInit () {inti,j;  for(i =0; I <3; i + +)         for(j =0; J <3; j + +) A[i][j]=0;}intJudge (intLen) {    inti,j;  for(i =2; I <= Len;i + +)         for(j = i+1; J <= Len;j + +){              if((m[i]-1)/3= = (m[i-1]-1)/3)                {                    if((m[i]-1)/3= = (m[j]-1)/3&& ((M[i]-m[j]) * (m[i-1]-M[J]) <0))                        return 0; }              Else              {                  if(((m[i]-1)/3-(m[j]-1)/3) * ((m[i-1]-1)/3-(m[j]/3-1)) <0)                   return 0; }        }        return 1;}intMain () {intCAs; Charm[ -]; CIN>>CAs;    GetChar ();  while(CAS--) {cin.Get(M, -);        GetChar (); //cout << m << Endl;Init (); intLen =strlen (m); //cout << "len=" << len << Endl; //cout << len << Endl;        intj =1;  for(intI=0; i < len;i + =2) {m[j+ +] = M[i]-'0'; }        intAns = Judge (len/2+1); if(ans) cout<<"valid"<<Endl; Elsecout<<"Invalid"<<Endl; }}

3.King ' s Order

#include <iostream>#defineJdz (a) a < 0?-A:Ausing namespacestd;Long Longdp[2001][ -],sum[2001];voidinit () {inti; sum[0] = -; sum[1] = -; sum[2] =676; sum[3] =17576;  for(i =0; I <= -; i + +) {dp[0][i] = -; dp[1][i] =1; dp[2][i] = -; dp[3][i] =676; }}/*A long long sum (int n) {Long long s = 1;    while (n--) {s = (S * 26)%1000000007; } return s;}*/intMain () {intCas,n; Long Longans; CIN>>CAs;  while(CAS--) {ans=0; CIN>>N; inti;        Init ();  for(inti =4; I <= N;i + +) {Sum[i]=0;  for(intj =1; J <= -; j + +) {Dp[i][j]= Jdz (( -*dp[i-1][j]%1000000007-sum[i-4]%1000000007+dp[i-4][j]%1000000007))%1000000007; Sum[i]= (sum[i]%1000000007+ dp[i][j]%1000000007)%1000000007; }        }         for(i =1; I <= -; i + +) ans= (ans%1000000007+ dp[n][i]%1000000007)%1000000007; cout<< ans <<Endl; }    return 0;}

Bestcoder Round #75 Problem Solving report