Codeforces Round #300

Http://codeforces.com/contest/538A. Cutting Banner

S=raw_input ()
flag = 0 for
i in range (len (s)):
    for J in range (I, Len (s)):
        if (s[:i]+s[j+1:] = = ' Codeforces '):
            flag = 1
if (flag==1):
    print ' YES '
else:
    print ' NO '

B. Quasi Binary

n = input ()
ans = [] while
n > 0:
    tmp = n
    i = 0
    y = 1
    res = 0 while
    tmp > 0:x = tmp%10
        tmp/=
        if x > 0:
            res + = y
        y *=
    N-= res
    ans.append (res)
print len ( ANS)
print ". Join (Map (str, ans))

C. Tourist ' s Notes

N, m = map (int, raw_input (). Split ())
PD, ph = map (int, raw_input (). Split ())
ans = ph+pd-1 for
i in range (m-1):
    d, h = map (int, raw_input (). Split ())
    xd = d-pd
    XH = ABS (H-PH)
    if XH > xd:
        print ' Impossible '
        E XIT ()
    tmp = (d-pd+h+ph)/2
    ans = max (ans, tmp)
    PD = d
    ph = h
ans = max (ans, ph+n-pd)
print ans

D. Weird chess Below is a python timeout code, changed to C + + can be ... No problem. Enumerate each position, determine whether to put ' x ', if relative to each o, there is no ' x ', then do not need, if one is ' x ', then need to turn him into ' X ', and finally judge whether it is legal ...

n = input()
mp = []
a = []
b = []
for i in range(n):
    s = raw_input()
    for j, ch in enumerate(s):
        if ch == 'o':
            a.append((i,j))
    mp.append(s)
ans = [['.' for i in range(2*n-1)] for i in range(2*n-1)]
ans2 = [['.' for i in range(n)] for i in range(n)]
tmp = 0
l = len(a)
for i in range(2*n-1):
    for j in range(2*n-1):
        sum1 = 0
        for oo in a:
            tx = oo[0]+(i+1-n)
            ty = oo[1]+(j+1-n)
            if tx<0 or tx>=n or ty<0 or ty>=n :
                sum1 += 1
                continue
            elif mp[tx][ty] == 'o' or mp[tx][ty] == 'x':
                sum1 = sum1+1
            else:
                break
        if sum1 == l:
            b.append((i,j))
            ans[i][j] = 'x'
for pp in b:
    for oo in a:
        tx = oo[0]+(pp[0]+1-n)
        ty = oo[1]+(pp[1]+1-n)
        if tx<0 or tx>=n or ty<0 or ty>=n :
            continue
        ans2[tx][ty] = 'x';
for i in range(n):
    for j in range(n):
        if mp[i][j] == 'o' or ans2[i][j] == 'o':
            continue
        if mp[i][j] != ans2[i][j]:
            tmp = 1
            break
if tmp:
    print 'NO'
else:
    ans[n-1][n-1] = 'o'
    print 'YES'
    for i in range(2*n-1):
        print ''.join(ans[i])

E. demiurges Play again tree DPF. A Heap of heaps matrix fast power