Chocolate&&木块拼接&&Cards

B. Chocolate

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob loves everything sweet. His favorite chocolate bar consists of pieces, each piece may contain a nut. Bob wants to break the bar of chocolate into multiple pieces so that each part would contain exactly one nut and any break line goes between two adjacent pieces.

You are asked to calculate the number of ways he can do it. Two ways to break chocolate are considered distinct if one of them contains a break between some two adjacent pieces and the other one doesn't.

Please note, that if Bob doesn't make any breaks, all the bar will form one piece and it still has to have exactly one nut.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of pieces in the chocolate bar.

The second line contains n integers a_i (0 ≤ a_i ≤ 1), where 0 represents a piece without the nut and 1 stands for a piece with the nut.

Output

Print the number of ways to break the chocolate into multiple parts so that each part would contain exactly one nut.

Examples

input

3 0 1 0

output

1

input

5 1 0 1 0 1

output

4
题解：插孔
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
int a[110];
int main(){
    int N;
    while(~scanf("%d",&N)){
        int t1=N,t2=N;
        long long ans=1;
        for(int i=0;i<N;i++){
            scanf("%d",&a[i]);
        }
        for(int i=0;i<N;i++){
            if(a[i]){
                t1=i;break;
            }
        }
        for(int i=N-1;i>=0;i--){
            if(a[i]){
                t2=i;break;
            }
        }
        if(t1==N){
            ans=1;
            printf("%d\n",0);continue;
        }
        int temp=1;
        for(int i=t1+1;i<=t2;i++){
            if(!a[i])temp++;
            else ans*=temp,temp=1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

 

• [1655] 木块拼接

• 时间限制: 1000 ms　内存限制: 65535 K
• 问题描述
• 好奇的skyv95想要做一个正方形的木块，现在有三种颜色的矩形木块，颜色分别为"A","B","C"。他现在很想把三个木块拼接成一个大正方形，现在求助于你们，问分别给你们三种颜色矩形的两个边长，判断是否能组成一个正方形。
• 输入
• 依次输入颜色为A的矩形的两边长度，颜色为B的矩形的两边长度，颜色为C的矩形的两边长度。
• 输出
• 可以输出"YES"，否则输出"NO"。
• 样例输入

• 4 4 2 6 4 2

• 样例输出

• YES

• 提示

• 例子的图像可以是这样
  6
  BBBBBB
  BBBBBB
  AAAACC
  AAAACC
  AAAACC
  AAAACC

• 题解：

• #include<iostream>
  #include<cstdio>
  #include<cstring>
  #include<cmath>
  #include<algorithm>
  using namespace std;
  const int INF=0x3f3f3f3f;
  #define mem(x,y) memset(x,y,sizeof(x))
  #define SI(x) scanf("%d",&x)
  #define PI(x) printf("%d",x)
  #define P_ printf(" ")
  struct Node{
      int x,y;
      bool operator < (const Node b) const{
          if(x!=b.x)return b.x<x;
          else return b.y<y;
      }
  };
  Node dt[3];
  bool js(int a,int b,int c,int d,int e,int f){
      //printf("%d %d %d %d %d %d\n",a,b,c,d,e,f);
      if(a==c+f&&b+d==b+e&&a==b+d)return true;
      if(a==d+e&&c==f&&a==b+d)return true;
      if(a==d+f&&c==e&&a==b+c)return true;
      if(a==c+e&&d==f&&a==b+d)return true;
      if(a==b+d+f&&a==c&&a==e)return true;
      return false;
  }
  int main(){
      int a[10];
      while(~scanf("%d%d%d%d%d%d",&dt[0].x,&dt[0].y,&dt[1].x,&dt[1].y,&dt[2].x,&dt[2].y)){
          for(int i=0;i<3;i++){
              if(dt[i].x<dt[i].y)swap(dt[i].x,dt[i].y);
          }
          sort(dt,dt+3);
          int flot=0;
          if(js(dt[0].x,dt[0].y,dt[1].x,dt[1].y,dt[2].x,dt[2].y))puts("YES");
          else puts("NO");
      }
      return 0;
  }

  B. Cards

  time limit per test

  2 seconds

  memory limit per test

  256 megabytes

  input

  standard input

  output

  standard output

  Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:

  • take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
  • take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.

  She repeats this process until there is only one card left. What are the possible colors for the final card?

  Input

  The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.

  The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.

  Output

  Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.

  Examples

  input

  2 RB

  output

  G

  input

  3 GRG

  output

  BR

  input

  5 BBBBB

  output

  B
  题解：规律
  代码：

  #include<iostream>
  #include<cstdio>
  #include<cstring>
  #include<cmath>
  #include<algorithm>
  using namespace std;
  const int INF=0x3f3f3f3f;
  #define mem(x,y) memset(x,y,sizeof(x))
  #define SI(x) scanf("%d",&x)
  #define PI(x) printf("%d",x)
  #define P_ printf(" ")
  const int MAXN=210;
  char s[MAXN];
  int g,r,b;
  int main(){
      int n;
      while(~scanf("%d",&n)){
          scanf("%s",s);
          g=r=b=0;
          for(int i=0;s[i];i++){
              if(s[i]=='R')r++;
              if(s[i]=='G')g++;
              if(s[i]=='B')b++;
          }
          if(r&&g&&b){
              puts("BGR");continue;
          }
          if(r+g==0||r+b==0||g+b==0){
              if(r)puts("R");
              else if(g)puts("G");
              else if(b)puts("B");
              else while(1);
              continue;
          }
          if(r+b+g==2){
              if(!b)puts("B");
              else if(!g)puts("G");
              else if(!r)puts("R");
              else while(1);
              continue;
          }
          if(r==1||b==1||g==1){
              if(r&&r!=1){
                  puts("BG");
              }
              else if(g&&g!=1)puts("BR");
              else if(b&&b!=1)puts("GR");
              else while(1);
              continue;
          }
          puts("BGR");
      }
      return 0;
  }