Codeforces Round #354 (Div. 2) B. Pyramid of Glasses (analog + thinking)

The original question, please poke here

Test instructions

Put the cup into a Yang Hui triangle, that is, the top 1 cups, the second Layer 2 cups, ... Nth Layer N Cups.

Every second can fill 1 cups, each time a cup is full, the cocktail party is divided into the bottom of it to support its two cups.

As shown in. 1 ≤ n ≤ 10, 0≤ t ≤10 000.

Analysis: Because n is small, it is a recursive process to directly simulate this process.

Note that if recursion is not recursive to the n+1 layer, then the last statistic is the number of statistics >=1 rather than the number of ==1,

Because when the wine can fill the cup larger than the N-Layer Cup (n+1)/2<t, the recursive process terminates on the N-level and cannot be pushed down. So the bottom

A layer of wine is greater than 1. The code is as follows:

#include <cstdio>#include<map>#include<algorithm>#include<iostream>#include<Set>#include<cmath>#include<cstring>using namespaceStd;typedefLong Longll;Doublep[ A][ A];intMain () {intn,t;  while(~SCANF ("%d%d",&n,&t) {memset (P,0,sizeof(p));  for(intI=1; i<=t;i++) {p[1][1]+=1;  for(intj=1; j<n;j++)            {                 for(intk=1; k<=j;k++)                {                    //cout<<j<< ' <<k<<endl;                    if(p[j][k]>1) {p[j+1][k] + = (p[j][k]-1)/2; P[j+1][k+1] + = (p[j][k]-1)/2; P[J][K]=1; //cout<<i<<endl; //cout<<j+1<< ' <<k<< ' <<p[j+1][k]<<endl; //cout<<j+1<< ' <<k+1<< ' <<p[j+1][k+1]<<endl;                    }                }            }        }        intCNT =0;  for(intj=1; j<=n;j++)        {             for(intk=1; k<=j;k++)            {                if(p[j][k]>=1) CNT++; }} cout<<cnt<<Endl; }    return 0;}

View Code

If you push to the n+1 layer, the number of ==1 is counted.

#include <cstdio>#include<map>#include<algorithm>#include<iostream>#include<Set>#include<cmath>#include<cstring>using namespaceStd;typedefLong Longll;Doublep[ A][ A];intMain () {intn,t;  while(~SCANF ("%d%d",&n,&t) {memset (P,0,sizeof(p));  for(intI=1; i<=t;i++) {p[1][1]+=1;  for(intj=1; j<=n;j++)//<=n guaranteed recursion to n+1 layer            {                 for(intk=1; k<=j;k++)                {                    if(p[j][k]>1) {p[j+1][k] + = (p[j][k]-1)/2; P[j+1][k+1] + = (p[j][k]-1)/2; P[J][K]=1; }                }            }        }        intCNT =0;  for(intj=1; j<=n;j++)        {             for(intk=1; k<=j;k++)            {                if(p[j][k]==1)//cnt++; }} cout<<cnt<<Endl; }    return 0;}

Codeforces Round #354 (Div. 2) B. Pyramid of Glasses (analog + thinking)