s = input().strip()
dp = [0] * (len(s) + 1)
dp[1] = ord(s[0]) - 96
for i in range(2,len(s) + 1):
dp[i] = max(dp[i - 2] + ord(s[i - 1]) - 96,dp[i - 1])
print(dp[len(s)])
#第I题：样例输入第一行为"4 4"
#二分答案 + 区间合并
def check(t):#判断t时刻 是否能让管道全部检测到水
brr = []#每个在t时刻 打开的阀门的左右区间 宽度
for i in range(n):
if arr[i][0] > t:break
Si,Li = arr[i]
#保持l最小是1 r最大是Len 超过就没有必要判断了
l,r = max(1,Li - (t - Si)),min(Len,Li + (t - Si))
brr.append((l,r))
brr.sort(key=lambda x:x[0])#按左端点排序
#区间合并
end = 0
for l,r in brr:
if l - end > 1:return False
end = max(end,r)
if end == Len:return True
return end == Len
def main():
l,r = 1,Len + arr[-1][0] + 100
ans = r
while l <= r:
mid = (l + r) // 2
if check(mid):ans = mid;r = mid - 1
else:l = mid + 1
return ans
n,Len = map(int,input().strip().split())
arr = list()#阀门a[i][0]-什么时刻打开 a[i][1] - 阀门编号
for i in range(n):
L,S = map(int,input().strip().split())
arr.append((S,L))
arr.sort(key=lambda x:x[0])#按打开时刻排序
ans = main()
print(ans)
#贪心 低位 -> 高位 遍历
#每次衡量当前位 到目标值 所需最小花费
#进位后，还要判断对进位后数字的影响
#返回x[u] 一直加1 到与y[u]位相同所需要的花费和加完后有改动的字符串
def count(u,t):
res = 0
while u != 0 and t != 0:
c = x[u]
sub = (c - y[u] + 10) % 10
jf = (y[u] - c + 10) % 10
after = min(sub,jf)#进位前需要的最小花费
c += t
if c == 10:c = 0;t = 1
elif c == -1:c = 0;t = -1
else:t = 0
sub = (c - y[u] + 10) % 10
jf = (y[u] - c + 10) % 10
now = min(sub,jf)#进位后需要的最小花费
res += now - after
u -= 1
return res
#加法
def Sum(u):
res = y[u] - x[u]
if res < 0:
#否则表示要进位
res = y[u] + 10 - x[u]
res += count(u - 1,1)
return res
#减法
def Mul(u):
res = x[u] - y[u]
if res < 0:
#否则表示要借位
res = x[u] + 10 - y[u]
res += count(u - 1,-1)
return res
def main():
ans = 0#要移动的次数
r = n - 1
while r >= 0:
if x[r] == y[r]:r -= 1;continue
SumCnt,MulCnt = Sum(r),Mul(r)
## print(SumCnt,MulCnt)
if SumCnt <= MulCnt:#做加法
ans += (y[r] - x[r] + 10) % 10
## print("Sum",r,ans,(y[r] - x[r] + 10) % 10)
if y[r] < x[r]:#说明进位了
t,i = 1,r - 1#向前进位
while i != -1 and t != 0:
if x[i] == 9:x[i] = 0;t = 1
else:x[i] += t;t = 0
i -= 1
else:#做减法
ans += (x[r] - y[r] + 10) % 10
## print("Mul",r,ans,(x[r] - y[r] + 10) % 10)
if y[r] > x[r]:#要借位
t,i = -1,r - 1
while i != -1 and t != 0:
if x[i] == 0:x[i] = 9;t = -1
else:x[i] += t;t = 0
i -= 1
## print(x[:r])
r -= 1
return ans
n = int(input().strip())
x,y = list(map(int,input().strip())),list(map(int,input().strip()))
##print(n,x,y)
ans = main()
print(ans)
我题意理解错了，我理解成选了第i层后，第i-1和第i+1层都不能选…
def add(a,b):
node.append((b,h[a]))
h[a] = len(node) - 1
def dfs(f,root):
#要么选
floot[f].append(root)
t = h[root]
while t != -1:
b,t = node[t]
dfs(f + 1,b)
def main():
for i in range(n,0,-1):
for j in floot[i]:
dp[i] = max(dp[i],v[j])
if i + 2 <= n:dp[i] += dp[i + 2]
n = int(input().strip())
node,h = [0],[-1] * (n + 1)#邻接表存储树
par = list(map(int,input().strip().split()))
v = list(map(int,('0 ' + input()).strip().split()))
for i in range(2,n + 1):
add(par[i - 2],i)
floot = [[] for i in range(n + 1)]#每一层的结点有哪些
dfs(1,1)#划分层级
dp = [0] * (n + 1)
main()
ans = dp[1]
if n >= 2:ans = max(ans,dp[2])
print(ans)
def add(x,y,z):
node.append((y,z,h[x]))
h[x] = len(node) - 1
def dijk(start):
dis[start] = 0
for i in range(n):
t = -1
for j in range(n):
if st[j]:continue
if t == -1 or dis[t] > dis[j]:
t = j
st[t] = True
j = h[t]
while j != -1:
y,z,j = node[j]
if st[y]:continue
if dis[y] == dis[t] + z:
path[y].append(t)
elif dis[y] > dis[t] + z:
dis[y] = dis[t] + z
path[y] = [t]#只能从t转移
def count(x):
res,st = 0,[False] * (n + 1)
Que = [x]
while len(Que) != 0:
t = Que.pop()
if st[t]:continue
for i in path[t]:
if i == 1:
res += 1
st[t] = True
else:
Que.append(i)
return res
n,m = map(int,input().strip().split())
node,h = [0],[-1] * (n + 1)
for i in range(m):
x,y,z = map(int,input().strip().split())
add(x,y,z)
add(y,x,z)
dis,st = [float('inf')] * (n + 1),[False] * (n + 1)
path = [[] for i in range(n + 1)]#i从哪个最短路过来的
dijk(1)
path[1] = [1]#1默认是从自己过来的
for i in range(1,n + 1):
if dis[i] == float('inf'):print(-1)
else:print(count(i) - 1)
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