一、第一题：教室

解题思路：二分+差分
        对天数进行二分，在ck函数中用差分方法优化多次区间累加。

【Python程序代码】

n,m = map(int,input().split())
a = [0] + list(map(int,input().split()))
d,s,t = [0]*(m+5),[0]*(m+5),[0]*(m+5)
for i in range(1,m+1):
    d[i],s[i],t[i] = map(int,input().split())
l,r = 1,m+1
def ck(mid):
    tep = [0]*(n+5)
    for i in range(1,mid+1):
        tep[s[i]] += d[i]
        tep[t[i]+1] -= d[i]
    for i in range(1,n+1):
        tep[i] += tep[i-1]
        if tep[i]>a[i]:return False
    return True
while l<r:
    mid = (l+r+1)//2
    if ck(mid):l = mid
    else:r = mid-1
    
if r>m:print(0)
else:
    print(-1)
    print(r)

---

 二、第二题：分巧克力

解题思路：二分
        对边长二分，所以每块大巧克力最多可以分成(h[i]/mid)*(w[i]/mid)。

【Python程序代码】

n,k = map(int,input().split())
h,w = [],[]
for i in range(n):
    a,b = map(int,input().split())
    h.append(a); w.append(b)
l,r = 1,100000
def ck(mid):
    res = 0
    for i in range(n):
        res += (h[i]//mid)*(w[i]//mid)
    if res>=k:return True
    return False
while l<r:
    mid = (l+r+1)//2
    if ck(mid):l = mid
    else:r = mid-1
print(r)

---

三、第三题： 管道

 解题思路：二分+区间合并
        对时间进行二分，然后在ck函数中使用区间合并进行优化。下面程序貌似没AC.

【Python程序代码】

n,L = map(int,input().split())
sl = [[0,0]for i in range(n)]
tp = [[0,0]for i in range(n)]
for i in range(n):
    sl[i][0],sl[i][1] = map(int,input().split())
def cmp(x):
    return (x[0],x[1])
sl.sort(key=cmp)
l,r = 1,1000000000
def ck(mid):
    mi,ma = 2,L-1
    for i in range(n):
        if mid>=sl[i][1]:
            tp[i][0] = sl[i][0] - (mid-sl[i][1])
            tp[i][1] = sl[i][0] + (mid-sl[i][1])
            if tp[i][0]<mi:mi=tp[i][0]
            if tp[i][1]>ma:ma=tp[i][1]
        else:
            tp[i][0]=tp[i][1]=0
    if mi>1 or ma<L:return False
    tp.sort(key=cmp)
    ll = 0
    for i in range(n):
        if tp[i][0]==tp[i][1]==0:continue
        else:
            if tp[i][0]>ll+1:return False
            ll = max(ll,tp[i][1])
    if ll>=L:
        return True
    return False

while l<r:
    mid = (l+r)//2
    if ck(mid):r=mid
    else:l=mid+1
print(r)

---

四、第四题： 技能升级

 解题思路：二分
        对攻击力大小进行二分，题的本质考的是多路归并，采用二分进行优化。

【Python程序代码】

n,m = map(int,input().split())
a,b = [],[]
for i in range(n):
    a_,b_ = map(int,input().split())
    a.append(a_); b.append(b_)
l,r = 0,10000000

def ck(mid):
    res = 0
    for i in range(n):
        if a[i]<mid:continue
        res += (a[i]-mid)//b[i] + 1
        if res>=m:return True
    return False
while l<r:
    mid = (l+r+1)>>1
    if ck(mid):l = mid
    else:r = mid-1
cnt,res=m,0
for i in range(n):
    if a[i]<r:continue
    t = (a[i]-r)//b[i] +1
    cnt -= t
    res += (2*a[i]-(t-1)*b[i])*t//2
print(res+cnt*r)

---

五、第五题： 冶炼金属

 解题思路：二分
        对V值进行二分，分别找左右边界。

【Python程序代码】

n = int(input())
a,b = [0]*(n+5),[0]*(n+5)
for i in range(n):
    a[i],b[i] = map(int,input().split())
def pd1(x):
    for i in range(n):
        t = a[i]//x
        if t>b[i]:
            return False
        if t<b[i]:
            return True
    return True
def pd2(x):
    for i in range(n):
        t = a[i]//x
        if t>b[i]:
            return True
        if t<b[i]:
            return False
    return True
def solve():
    l,r = 1,1000000000
    while l<r:
        mid = (l+r)//2
        if pd1(mid):r=mid
        else:l=mid+1
    res1 = r
    l, r = 1, 1000000000
    while l < r:
        mid = (l + r + 1) // 2
        if pd2(mid):
            l = mid
        else:
            r = mid - 1
    res2 = r
    print(res1,res2)
solve()