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Description
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn‘t need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input
The first line of the input contains an integer n (1?≤?n?≤?100000) — the number of magnets. Thenn lines follow. The i-th line (1?≤?i?≤?n) contains either characters "01", if Mike put thei-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output
On the single line of the output print the number of groups of magnets.
Sample Input
6 10 10 10 01 10 10
3
4 01 01 10 10
2
Hint
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
/***************************************************************************************
Question:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121431#problem/A
Date:2016.07.10
Author:Liusiyu /
\ \ \ | | | | |
* \ \ \ \ \ \ __\--|__
** / / / / / / II=======OOOOO[/ ★02 __|
** / / | | | | _\______|/--*
* / | | | | | /_____________|
/ | | | | | | \◎◎◎◎◎◎◎◎⊙/
****************************************************************************************/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <vector>
#include <ctime>
#include <list>
#include <deque>
#include <bitset>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)<(b)?(a):(b)
#define FL(s,n,a) fill(s,s+n,a);
#define MAPIT(a,b) map<a,b>::iterator it;
#define VECIT(type) vector<type>::iterator it;
#define SETIT(type) set<type>::iterator it;
using namespace std;
typedef long long ll;
const int MAXN=1005;
//const int MINN=;
//const int INF=0x3f3f3f3f;
int main()
{
int N;
int sum;
char a[10],b[10];
while(scanf("%d",&N)!=EOF)
{
scanf("%s",a);
sum=0;
for(int i=1;i<N;i++)
{
scanf("%s",b);
if(b[0]==a[1])
sum++;
a[0]=b[0];
a[1]=b[1];
}
}
cout<<sum+1<<endl;
}
Mad scientist Mike is busy carrying out experiments in chemistry. Today he will attempt to join three atoms into one molecule.
A molecule consists of atoms, with some pairs of atoms connected by atomic bonds. Each atom has a valence number — the number of bonds the atom must form with other atoms. An atom can formone or multiple bonds with any other atom, but it cannot form a bond with itself. The number of bonds of an atom in the molecule must be equal to its valence number.
Mike knows valence numbers of the three atoms. Find a molecule that can be built from these atoms according to the stated rules, or determine that it is impossible.
Input
The single line of the input contains three space-separated integers a, b and c (1?≤?a,?b,?c?≤?106) — the valence numbers of the given atoms.
Output
If such a molecule can be built, print three space-separated integers — the number of bonds between the 1-st and the 2-nd, the 2-nd and the 3-rd, the 3-rd and the 1-st atoms, correspondingly. If there are multiple solutions, output any of them. If there is no solution, print "Impossible" (without the quotes).
Sample Input
1 1 2
0 1 1
3 4 5
1 3 2
4 1 1
Impossible
Hint
The first sample corresponds to the first figure. There are no bonds between atoms 1 and 2 in this case.
The second sample corresponds to the second figure. There is one or more bonds between each pair of atoms.
The third sample corresponds to the third figure. There is no solution, because an atom cannot form bonds with itself.
The configuration in the fourth figure is impossible as each atom must have at least one atomic bond.
/***************************************************************************************
Question:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121431#problem/B
Date:2016.07.10
Author:Liusiyu /
\ \ \ | | | | |
* \ \ \ \ \ \ __\--|__
** / / / / / / II=======OOOOO[/ ★02 __|
** / / | | | | _\______|/--*
* / | | | | | /_____________|
/ | | | | | | \◎◎◎◎◎◎◎◎⊙/
****************************************************************************************/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <vector>
#include <ctime>
#include <list>
#include <deque>
#include <bitset>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)<(b)?(a):(b)
#define FL(s,n,a) fill(s,s+n,a);
#define MAPIT(a,b) map<a,b>::iterator it;
#define VECIT(type) vector<type>::iterator it;
#define SETIT(type) set<type>::iterator it;
using namespace std;
typedef long long ll;
const int MAXN=1005;
//const int MINN=;
//const int INF=0x3f3f3f3f;
int main()
{
int A,B,C,x,y,z;
while(scanf("%d %d %d",&A,&B,&C)!=EOF)
{
x=A+B-C;
y=B-A+C;
if(x%2!=0||y%2!=0||y<0)
printf("Impossible");
else
{
z=2*C-y;
x=2*A-z;
if(x<0||z<0)
printf("Impossible");
else
printf("%d %d %d\n",x/2,y/2,z/2);
}
}
}
Description
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance R0?=?1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
With the consecutive connection the resistance of the new element equals
R?=?Re?+?R0. With the parallel connection the resistance of the new element equals.
In this caseRe equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine
the smallest possible number of resistors he needs to make such an element.
Input
The single input line contains two space-separated integers
a and b (1?≤?a,?b?≤?1018). It is guaranteed that the fraction
is irreducible. It is guaranteed that a solution always exists.
Output
Print a single number — the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use thecin, cout streams or the%I64d specifier.
Sample Input
1 1
1
3 2
3
199 200
200
Hint
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance.
We cannot make this element using two resistors.
/***************************************************************************************
Question:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121431#problem/B
Date:2016.07.10
Author:Liusiyu /
\ \ \ | | | | |
* \ \ \ \ \ \ __\--|__
** / / / / / / II=======OOOOO[/ ★02 __|
** / / | | | | _\______|/--*
* / | | | | | /_____________|
/ | | | | | | \◎◎◎◎◎◎◎◎⊙/
****************************************************************************************/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <vector>
#include <ctime>
#include <list>
#include <deque>
#include <bitset>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)<(b)?(a):(b)
#define FL(s,n,a) fill(s,s+n,a);
#define MAPIT(a,b) map<a,b>::iterator it;
#define VECIT(type) vector<type>::iterator it;
#define SETIT(type) set<type>::iterator it;
using namespace std;
typedef long long ll;
const int MAXN=1005;
//const int MINN=;
//const int INF=0x3f3f3f3f;
ll sum=0;
void deal(ll a,ll b)
{
if(a>b)
{
sum+=a/b;
a%=b;
}
if(a==1)
{
sum+=b;return;
}
else
{
sum+=b/a;
if(b%a==0)
return;
deal(b%a,a);
}
}
int main()
{
ll a,b;
scanf("%I64d %I64d",&a,&b);
//cout<<"a="<<a<<" b="<<b<<endl;
if(a%b==0)
{
printf("%I64d\n",a/b);
return 0;
}
deal(a,b);
printf("%I64d\n",sum);
}
Description
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them andwithout moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input
The single line of the input contains a sequence of characters "+" and "-" of lengthn (1?≤?n?≤?100000). Thei-th (1?≤?i?≤?n) position of the sequence contains the character "+", if on thei-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Sample Input
-++-
Yes
+-
No
++
Yes
-
No
Hint
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
/*************************************************************************
File Name: 1.cpp
ID: Liusiyu
PROG:
LANG: C++
Mail: 779532360@qq.com
Created Time: 2016年07月12日 星期二 09时33分16秒
************************************************************************/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <set>
#include <bitset>
#include <deque>
#include <vector>
#include <map>
#include <list>
#include <cstdlib>
#include <cassert>
#include <functional>
#include <iomanip>
#include <utility>
#include <complex>
#include <numeric>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
#define MAX(a,b) (a)>(b)?(a):(b);
#define MIN(a,b) (a)<(b)?(a):(b)
#define APS(a) (a)>0?(a):-(a)
#define FL(s,n,a) fill(s,s+n,a);
#define MAPIT(a,b) map<a,b>::iterator it;
#define VECIT(type) vector<type>::iterator it;
#define SETIT(type) set<type>::iterator it;
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
//const int MINN=;
const int INF=0x3f3f3f3f;
int main()
{
char a[MAXN];
char s[MAXN];
while(scanf("%s",a)!=EOF){
int len=strlen(a);
int flag=0;
for(int i=0;i<len;i++){
if(!flag||s[flag]!=a[i]){
flag++;
s[flag]=a[i];
}
else
flag--;
}
if(!flag)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
}
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原文地址：http://blog.csdn.net/r_o_j/article/details/51886587
