链接: http://codeforces.com/contest/1060
Codeforces Round #513
- A. Phone Numbers(水题)
- B. Maximum Sum of Digits(打表找规律、思维贪心)
- C. Maximum Subrectangle(思维卡时间)
A. Phone Numbers(水题)
题意: 第一位是字符
′
8
′
'8'
′8′ 并且长度正好
11
11
11 位的字符串是合法的。给你一个字符串,问有多少个合法的子串(子串可以相同)。
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[])
{
int n;
string s;
cin>>n>>s;
if(n < 11) {
cout<<0<<endl;
return 0;
}
int cnt = 0;
for (int i = 0; i < n; ++i) {
if(s[i] == '8') cnt++;
}
cout<<min(cnt, n/11)<<endl;
return 0;
}
B. Maximum Sum of Digits(打表找规律、思维贪心)
题意: 定义一个函数
S
(
x
)
S(x)
S(x) , 它的值为
x
x
x 的每一位数相加,比如
S
(
123
)
=
1
+
2
+
3
S(123) = 1+2+3
S(123)=1+2+3 。现在给你一个数
n
(
1
≤
n
≤
1
0
12
)
n (1 \leq n \leq 10^{12})
n(1≤n≤1012), 求最大的
S
(
a
)
+
S
(
b
)
(
a
+
b
=
n
,
0
≤
a
,
b
≤
n
)
S(a) + S(b) \ \ (a+b=n \ ,0 \leq a,b \leq n)
S(a)+S(b) (a+b=n ,0≤a,b≤n) 。
思路:
思路1 : 思考后可以发现,两个数
a
,
b
a,b
a,b,因为和为定值,
a
a
a 加
1
1
1,
b
b
b 减
1
1
1,在没有进位、退位的情况下,答案是不变的,直到
a
a
a 变成全是
9
9
9 这种数(比如
9999999
9999999
9999999),进位之后答案变小了,所以分解成
(
99999999
)
(99999999)
(99999999) 一定是最优的 。
思路2: 还有一种思路,打表观察发现
t
t
t 个
9
9
9 的这种数(比如
9999999
9999999
9999999), 最大值一定是
t
∗
9
t*9
t∗9,那么我们假设就尽可能的分出
9
9
9 ,这样分解最大。验证后确实如此。
打表代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main(int argc, char const *argv[])
{
for (LL n = 1; n <= 10000; ++n) {
LL ans = -1;
for (LL i = 0; i <= (n+1)/2; ++i) {
LL tmp = dig(i) + dig(n-i);
ans = max(ans, tmp);
}
cout<<ans<<endl;
}
return 0;
}
AC代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL dig(LL x) {
LL res = 0;
while(x) {
res += x%10;
x /= 10;
}
return res;
}
int main(int argc, char const *argv[])
{
LL n;
cin>>n;
if(n < 10) {
cout<<n<<endl;
return 0;
}
LL tmp = n;
LL cnt = 0;
while(tmp) {
cnt++;
tmp /= 10;
}
tmp = 0;
for (int i = 0; i < cnt-1; ++i) {
tmp *= 10;
tmp += 9;
}
LL ans = dig(tmp) + dig(n-tmp);
cout<<ans<<endl;
return 0;
}
C. Maximum Subrectangle(思维卡时间)
题意 : 给出数组
a
,
b
a, b
a,b ,定义矩阵
c
c
c,
c
[
i
]
[
j
]
=
a
[
i
]
∗
b
[
j
]
c[i][j] = a[i] * b[j]
c[i][j]=a[i]∗b[j]。现在想让你找一个子矩阵
(
1
≤
x
1
≤
x
2
≤
n
,
1
≤
y
1
≤
y
2
≤
m
)
(1 \leq x1 \leq x2 \leq n\ , 1 \leq y1 \leq y2 \leq m)
(1≤x1≤x2≤n ,1≤y1≤y2≤m)使得
∑
i
=
x
1
x
2
∑
j
=
y
1
y
2
c
[
i
]
[
j
]
≤
x
\sum _{i=x1} ^{x2} \sum _{j=y1} ^{y2} c[i][j] \leq x
∑i=x1x2∑j=y1y2c[i][j]≤x, 求最大的
s
=
(
y
2
−
y
1
+
1
)
∗
(
x
2
−
x
1
+
1
)
s = (y2-y1+1) * (x2-x1+1)
s=(y2−y1+1)∗(x2−x1+1)。
思路:
∑
i
=
x
1
x
2
∑
j
=
y
1
y
2
c
[
i
]
[
j
]
=
∑
i
=
x
1
x
2
∑
j
=
y
1
y
2
a
[
i
]
∗
b
[
j
]
=
(
∑
i
=
x
1
x
2
a
[
i
]
)
∗
(
∑
j
=
y
1
y
2
b
[
j
]
)
\sum _{i=x1} ^{x2} \sum _{j=y1} ^{y2} c[i][j] = \sum _{i=x1} ^{x2} \sum _{j=y1} ^{y2} a[i]*b[j] = (\sum _{i=x1} ^{x2} a[i])* (\sum _{j=y1} ^{y2} b[j])
i=x1∑x2j=y1∑y2c[i][j]=i=x1∑x2j=y1∑y2a[i]∗b[j]=(i=x1∑x2a[i])∗(j=y1∑y2b[j])
矩阵的和就是横向所有数的和乘以纵向所有数的和。
思路1:
预处理一个方向(比如横向)的长度为
i
i
i 的最小子段和,然后排序。暴力另一个方向(纵向)所有子段,二分找合法的最大横向子段和。复杂度
O
(
m
a
x
(
m
2
l
o
g
n
,
n
2
)
)
O(max(m^2 logn, n^2))
O(max(m2logn,n2)) 。
思路2:
分别预处理
a
,
b
a, b
a,b 的长度为
i
i
i 的最小子段和。然后分别枚举长度。复杂度
O
(
m
a
x
(
n
∗
m
,
n
2
,
m
2
)
)
O(max(n*m, n^2, m^2))
O(max(n∗m,n2,m2))。
个人推荐思路2
思
路
1
A
C
代
码
思路1AC代码
思路1AC代码
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MAXN = 2100;
const int INF = 0x3f3f3f3f;
LL lena[MAXN*MAXN];
struct node {
LL val;
LL num;
friend bool operator < ( node a, node b ) {
return a.val<b.val;
}
};
int main(int argc, char const *argv[])
{
LL n, m, a[MAXN], b[MAXN], x;
cin>>n>>m;
for (LL i = 1; i <= n; ++i) {
cin>>a[i];
}
for (LL i = 1; i <= m; ++i) {
cin>>b[i];
}
cin>>x;
LL suma[MAXN], sumb[MAXN];
suma[0] = 0;
for (LL i = 1; i <= n; ++i) {
suma[i] = suma[i-1] + a[i];
}
sumb[0] = 0;
for (LL i = 1; i <= m; ++i) {
sumb[i] = sumb[i-1] + b[i];
}
std::vector<node > sumaz;
memset(lena, INF, sizeof(lena));
for (LL i = 1; i <= n; ++i) {
for (LL j = i; j <= n; ++j) {
lena[j-i+1] = min(lena[j-i+1], suma[j] - suma[i-1]);
}
}
for (LL i = 1; i <= n; ++i) {
sumaz.push_back((node){lena[i], i});
}
sort(sumaz.begin(), sumaz.end());
LL s = 0;
for (LL i = 1; i <= m; ++i) {
for (LL j = i; j <= m; ++j) {
node tmp;
tmp.val = x/(sumb[j] - sumb[i-1]);
vector<node >::iterator it = lower_bound(sumaz.begin(), sumaz.end(), tmp);
if(it != sumaz.end()) {
if(((*it).val)*(sumb[j] - sumb[i-1]) <= x)
s = max(s, (j-i+1)*((*it).num));
else {
if(it != sumaz.begin()){
it--;
s = max(s, (j-i+1)*((*it).num));
}
}
}
else {
it--;
s = max(s, (j-i+1)*((*it).num));
}
}
}
cout<<s<<endl;
return 0;
}
思
路
2
A
C
代
码
思路2AC代码
思路2AC代码
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2100;
const int INF = 0x3f3f3f3f;
LL lena[MAXN*MAXN];
LL lenb[MAXN*MAXN];
int main(int argc, char const *argv[])
{
LL n, m, a[MAXN], b[MAXN], x;
cin>>n>>m;
for (LL i = 1; i <= n; ++i) {
cin>>a[i];
}
for (LL i = 1; i <= m; ++i) {
cin>>b[i];
}
cin>>x;
LL suma[MAXN], sumb[MAXN];
suma[0] = 0;
for (LL i = 1; i <= n; ++i) {
suma[i] = suma[i-1] + a[i];
}
sumb[0] = 0;
for (LL i = 1; i <= m; ++i) {
sumb[i] = sumb[i-1] + b[i];
}
memset(lena, INF, sizeof(lena));
for (LL i = 1; i <= n; ++i) {
for (LL j = i; j <= n; ++j) {
lena[j-i+1] = min(lena[j-i+1], suma[j] - suma[i-1]);
}
}
memset(lenb, INF, sizeof(lenb));
for (LL i = 1; i <= m; ++i) {
for (LL j = i; j <= m; ++j) {
lenb[j-i+1] = min(lenb[j-i+1], sumb[j] - sumb[i-1]);
}
}
LL s = 0;
for (LL i = 1; i <= n; ++i) {
for (LL j = 1; j <= m; ++j) {
if(lena[i] * lenb[j] <= x) {
s = max(s, i*j);
}
}
}
cout<<s<<endl;
return 0;
}
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