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.
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 ai (0 ≤ ai ≤ 1), where 0 represents a piece without the nut and 1 stands for a piece with the nut.
Print the number of ways to break the chocolate into multiple parts so that each part would contain exactly one nut.
3 0 1 0
1
5 1 0 1 0 1
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 ",0);continue; } int temp=1; for(int i=t1+1;i<=t2;i++){ if(!a[i])temp++; else ans*=temp,temp=1; } printf("%lld ",ans); } return 0; }
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[1655] 木块拼接
- 时间限制: 1000 ms 内存限制: 65535 K
- 问题描述
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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 ",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. Cardstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputCatherine 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?
InputThe 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.
OutputPrint 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.
Examplesinput2 RB
outputG
input3 GRG
outputBR
input5 BBBBB
outputB
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; }
直接记录x+y,x-y的数量，n*(n-1)/2和就是答案了；以前做过。。。
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(" ") typedef __int64 LL; const int MAXN=4200; int vis[MAXN]; int main(){ int N; while(~scanf("%d",&N)){ int x,y; mem(vis,0); while(N--){ scanf("%d%d",&x,&y); vis[x+y+2100]++; vis[x-y+1010]++; } LL ans=0; for(int i=0;i<MAXN;i++){ if(vis[i]){ ans+=(vis[i]-1)*vis[i]/2; } } printf("%I64d ",ans); } return 0; }