UVA232：题目就不写了,主要是模拟单词的变换.注意必须要按照启示格的顺序来输出.在输出竖的单词时候要用一个矩阵来标记是否被访问过.当然还可以按照竖的方式写入然后再排序.
def cross(matrix,r,c):
mark = [[0]*c for i in range(r)]
def test(i,j):
if(i==0 or j==0):return True
elif(matrix[i-1][j] == '*' or matrix[i][j-1]=='*'):return True
else:return False
i = 0;across = ''
while(i<r):
j = 0
while(j<c):
if(matrix[i][j]!='*' and test(i,j)):
across = ''
across+=matrix[i][j]
while(j+1<c and matrix[i][j+1]!='*'):across+=matrix[i][j+1];j+=1
j+=1
print(across)
else: j+=1
i+=1
print('Down:')
i = 0;across = ''
while(i<r):
j = 0
while(j<c):
if(matrix[i][j]!='*' and test(i,j) and mark[i][j]!=1):
across = ''
mark[i][j] = 1
across+=matrix[i][j]
_i = i;
while(_i+1<r and matrix[_i+1][j]!='*'):
across+=matrix[_i+1][j]
mark[_i+1][j] = 1
_i+=1
j+=1
print(across)
else: j+=1
i+=1
UVA1368:DNA序列
只需要一列一列的进行比较就可以了,因为每一个位置的DNA都是独立的,与其它DNA无关
def Hamming(buf,m,n):
ans = ''
diff = 0
for j in range(n):
count = {'A':0,'C':0,'G':0,'T':0}
max = 0
for i in range(m):
count[buf[i][j]]+=1
if(count[buf[i][j]] > max):
max = count[buf[i][j]]
max_alpha = buf[i][j]
elif(count[buf[i][j]] == max):
if(buf[i][j]<max_alpha):
max = count[buf[i][j]]
max_alpha = buf[i][j]
ans += max_alpha
diff += (m - max)
print(ans,diff)
Hamming(['TATGATAC','TAAGCTAC','AAAGATCC','TGAGATAC','TAAGATGT'],5,8)
UVA202，这个题目的关键不是去判断小数,而是判断每一次的余数,只要余数重复那么小数就会重复的.想到这一点就不难了.
def Repeating(a,b):
ans1 = a//b
a = a%b
tmp = ''
ans2 = str((a*10)//b)
mark_store = []
while(True):
a = (a*10)%b
tmp = (a*10)//b
if(a in mark_store):break
else:
mark_store.append(a)
ans2 += str(tmp)
m = ans2[:-1]
print(str(ans1)+'.'+'(%s)'%(m),len(m))
UVA10340：这个题目关键是要逆着思考,考虑s字符串中的字符是否可以在t中按顺序找到,即只能向后查找.这样就能很容易的判断了.O(m+n)的复杂度。
def Allinall(s,t):
index = -1
for ch in s:
index = t.find(ch,index+1)
if(index == -1):print('NO');return False
print('YES')
Allinall('dc','abcde')
UVA1587,此题关键是想到将输入标准化,统计边的情况之后进行检验.一定满足3种的分布.3条不同的边,2条相同,3条都一样.
def Box(L,W):
def mysort():
for i in range(6):
if(L[i]<W[i]):L[i],W[i] = W[i],L[i]
mysort()
X = [(L[i],W[i]) for i in range(6)]
Count = {}
for each in X:
if str(each) not in Count:
Count[str(each)] = 1
else:Count[str(each)] += 1
if(len(Count)==3):
print(Count)
for each,val in Count.items():
if(val != 2):print('Impossible 4');return
print('Possible')
elif(len(Count)==2):
test = [x for k,x in Count.items()]
test.sort()
if (test!=[2,4]):print('Impossible 8');return
print('Possible')
elif(len(Count)==1):
for each,val in Count.items():
if(Count[each]!=6):print('Impossible 12');return
print('Possible')
else:print('Impossible');return
Box([3,3,4,4,4,4],[3,3,5,3,3,3])
UVA1588,有个陷阱要注意,注意有两种模拟的方向.直接模拟,先放入长一点的长条,然后从长的长条每一个点开始遍历.然后放入短的长条进行遍历,最后取最小值
def Kickdown(h1,h2):
n1 = len(h1);n2 = len(h2)
def geth(i):
if(i>=n1):return 0
else: return h1[i]
def geth2(i):
if(i>=n2):return 0
else: return h2[i]
for i in range(n1+1):
if(geth(i)+h2[0]<=3):
test = True
for j in range(1,n2):
if(geth(i+j)+h2[j]>3):test = False;break
if(test):ans1 = max(n2+i,n1);break
for i in range(n2+1):
if(h1[0]+geth2(i)<=3):
test = True
for j in range(1,n1):
if(h1[j]+geth2(i+j)>3):test = False;break
if(test):ans2 = max(n1+i,n2);break
print(min(ans1,ans2))
