Codeforces Round #359 (Div. 2) C. Robbers ' Watch

Topic Link: Portal

The main topic: there is a table, by 7, you will be divided into n hours a day, an hour into M minutes, the table is required to show the number of different, ask how many legal time in the n,m limit

For example n=2,m=3 (0:1), (0:2), ( 1:0), ( 1:2). n=8,m=2 (02:1), (03:1), ( 04:1), ( 05:1 ), (06:1).

Topic idea: Because it is represented by 7, so the number displayed on the watch can not be more than 7, otherwise there must be duplicate (illegal), so we only need a special sentence, and then Dfs can.

This question is still more regrettable, at that time did not make out at night, woke up in the morning to find the boundary situation a bit of a problem, changed a bit over, oneself or is too weak.

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<cstring>#include<stack>#include<cctype>#include<queue>#include<string>#include<vector>#include<Set>#include<map>#include<climits>#defineLson Root<<1,l,mid#defineRson Root<<1|1,mid+1,r#defineFi first#defineSe Second#definePing (x, y) ((x-y) * (x-y))#defineMST (x, y) memset (x,y,sizeof (x))#defineMCP (x, y) memcpy (x,y,sizeof (y))using namespacestd;#defineGamma 0.5772156649015328606065120#defineMOD 1000000007#defineINF 0x3f3f3f3f#defineN 100005#defineMAXN 1005typedef pair<int,int>Pii;typedefLong LongLL;intn,m,k,x,num,y;intvis[Ten],ans;voidDFS2 (intSumintCNT) {///dfs2 processing minutes (m), other and DFS1 empathy    if(sum>=m)return; if(cnt==y) {++ans;return;}  for(intI=0;i<7;++i) {        if(!Vis[i]) {Vis[i]=1; Sum=sum*7+i; DFS2 (sum,cnt+1); Sum= (sum-i)/7; Vis[i]=0; }    }}voidDFS1 (intSumintCNT) {///DFS1 Processing hour segment (n), sum represents the current time (sum<n),if (sum>=n) return; ///CNT says several numbers have been processed.    if(cnt==x) {         for(intj=0;j<7; ++j)if(!Vis[j]) {Vis[j]=1; DFS2 (J,1); VIS[J]=0; }    }     for(intI=0;i<7;++i) {        if(!Vis[i]) {Vis[i]=1; Sum=sum*7+i; DFS1 (sum,cnt+1); Sum= (sum-i)/7; Vis[i]=0; }    }}intDealintx) {///Special handling, returns how many digits are required to represent n,m    intCnt=0;  while(x) {++CNT; X=x/7; }    returnCNT;}intMain () {inti,j,group,v; scanf ("%d%d",&n,&m); if(n%7==0) X=deal (n1);///It is important to note that if n can be divisible by 7    ElseX=deal (n);///so the number of digits required for n is reduced by one, that is, it is not considered clear, it is a pity.     if(m%7==0) Y=deal (M-1);///m-Empathy processing    Elsey=deal (m); if(x+y>7) {exit (0*printf ("0\n"));}  for(i=0;i<7;++i) {Vis[i]=1; DFS1 (i,1); Vis[i]=0; } printf ("%d\n", ans); return 0;}

Codeforces Round #359 (Div. 2) C. Robbers ' Watch