A.
暴力计算即可。
-
// #pragma GCC optimize(2)
-
#include <iostream>
-
#include <cstdio>
-
#include <algorithm>
-
#include <queue>
-
#include <cmath>
-
#include <string>
-
#include <vector>
-
#include <stack>
-
#include <map>
-
#include <sstream>
-
#include <cstring>
-
#include <set>
-
#include <cctype>
-
#include <bitset>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
2e8 +
10;
-
const
int maxm =
2e5 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
4][
2] = {
1,
0,
0,
-1,
0,
1,
-1,
0};
-
int m[
13] = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31};
-
int ans =
0;
-
int count(int n)
-
{
-
int cnt =
0;
-
while (n)
-
{
-
int t = n %
10;
-
if (t ==
2)
-
cnt++;
-
n = n /
10;
-
}
-
return cnt;
-
}
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
IO;
-
for (
int i =
1; i <=
2020; i++)
-
ans += count(i);
-
cout << ans;
-
return
0;
-
}
624
容易找规律发现相邻两项之间的关系。
-
// #pragma GCC optimize(2)
-
#include <iostream>
-
#include <cstdio>
-
#include <algorithm>
-
#include <queue>
-
#include <cmath>
-
#include <string>
-
#include <vector>
-
#include <stack>
-
#include <map>
-
#include <sstream>
-
#include <cstring>
-
#include <set>
-
#include <cctype>
-
#include <bitset>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
2e8 +
10;
-
const
int maxm =
2e5 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
4][
2] = {
1,
0,
0,
-1,
0,
1,
-1,
0};
-
int m[
13] = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31};
-
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
IO;
-
int a[
50];
-
int t =
1;
-
a[
1] =
1;
-
for (
int i =
2; i <=
20;i++)
-
a[i] = a[i -
1] + t *
4, t++;
-
cout << a[
20];
-
return
0;
-
}
761
C.
暴力判断即可。
-
// #pragma GCC optimize(2)
-
#include <iostream>
-
#include <cstdio>
-
#include <algorithm>
-
#include <queue>
-
#include <cmath>
-
#include <string>
-
#include <vector>
-
#include <stack>
-
#include <map>
-
#include <sstream>
-
#include <cstring>
-
#include <set>
-
#include <cctype>
-
#include <bitset>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
2e8 +
10;
-
const
int maxm =
2e5 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
4][
2] = {
1,
0,
0,
-1,
0,
1,
-1,
0};
-
int m[
13] = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31};
-
int gcd(int a, int b)
-
{
-
if (b ==
0)
-
return a;
-
else
-
return gcd(b, a % b);
-
}
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
IO;
-
int ans =
0;
-
for (
int i =
1; i <=
2020; i++)
-
for (
int j =
1; j <=
2020; j++)
-
if (gcd(i, j) ==
1)
-
ans++;
-
cout << ans;
-
return
0;
-
}
2481215
D.
比赛的时候因为怕数迷，就没用日历表数,现在回想一下真该检查的时候数一遍。
这题我写的8880，就是因为打印函数返回值出的锅（调用函数会导致全局变量改变），但凡当时用变量存一下函数返回值也不会错了，一念之间。。。省一没了。。。。
-
// #pragma GCC optimize(2)
-
#include <iostream>
-
#include <cstdio>
-
#include <algorithm>
-
#include <queue>
-
#include <cmath>
-
#include <string>
-
#include <vector>
-
#include <stack>
-
#include <map>
-
#include <sstream>
-
#include <cstring>
-
#include <set>
-
#include <cctype>
-
#include <bitset>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
2e8 +
10;
-
const
int maxm =
2e5 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
4][
2] = {
1,
0,
0,
-1,
0,
1,
-1,
0};
-
int m[
13] = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31};
-
int w =
5;
-
bool rui(int n)
-
{
-
if (n %
400 ==
0 || (n %
4 ==
0 && n %
100 !=
0))
-
return
true;
-
return
false;
-
}
-
int count(int year)
-
{
-
int cnt =
0;
-
if (rui(year))
-
m[
2] +=
1;
-
if (year ==
2020)
-
{
-
for (
int i =
1; i <=
9; i++)
-
{
-
for (
int j =
1; j <= m[i]; j++)
-
{
-
if (w ==
0 || j ==
1)
-
{
-
cnt++;
-
}
-
w = (w +
1) %
7;
-
}
-
}
-
}
-
else
-
{
-
for (
int i =
1; i <=
12; i++)
-
{
-
for (
int j =
1; j <= m[i]; j++)
-
{
-
if (w ==
0 || j ==
1)
-
{
-
cnt++;
-
}
-
w = (w +
1) %
7;
-
}
-
}
-
}
-
if (rui(year))
-
m[
2] -=
1;
-
return cnt;
-
}
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
IO;
-
int ans =
0;
-
for (
int i =
2000; i <=
2020; i++)
-
{
-
// cout << count(i) << endl;
-
// 一念之间 天堂地狱
-
// 闲的没事非打印干嘛？！
-
// 打印完了还不注释掉？！
-
ans += count(i);
-
}
-
cout << ans +
7580 +
1;
// 最后再加上10.1这一天
-
return
0;
-
}
8879
E.
可以用二维数组建图然后跑bfs判断是否连通。
-
#include<bits/stdc++.h>
-
using
namespace
std;
-
int
dis[6][2]
= {
1,
-1,
-1,
1,
-1,
-1,
1,
1,
2,
0,
-2,
0}
;
-
struct
Node
-
{
-
int
x,
y;
-
Node(){}
;
-
Node(int
a,
int
b)
-
{
-
x
=
a,
y
=
b;
-
}
-
}
;
-
int
g[10][10];
-
bool
book[10][10];
-
bool
vis[10];
-
int
a[10];
-
int
C,
ans;
-
bool
BFS(int
all)
-
{
-
memset(book,
false,
sizeof
book);
-
int
sx,
sy;
-
for
(int
i
=
1
;
i
<=
5
;
i++)
-
for
(int
j
=
1
;
j
<=
3
;
j++)
-
if
(g[
i][
j]
==
1
)
-
{
-
sx
=
i,
sy
=
j;
-
break;
-
}
-
-
queue<Node>
q;
-
q.push(Node(sx,
sy));
-
book[
sx][
sy]
=
true
;
-
int
cnt
=
0
;
-
while
(!q.empty())
-
{
-
Node
now
=
q.front();
-
q.pop();
-
cnt++;
-
if
(now.y
==
2
)
-
{
-
for
(int
i
=
0
;
i
<
4
;
i++)
-
{
-
int
tx
=
now.x
+
dis[
i][
0]
;
-
int
ty
=
now.y
+
dis[
i][
1]
;
-
if
(tx
>=
1
&&
tx
<=
5
&&
ty
>=
1
&&
ty
<=
3
&&
g[
tx][
ty]
==
1
&&
book[
tx][
ty]
==
false
)
-
{
-
book[
tx][
ty]
=
true
;
-
q.push(Node(tx,
ty));
-
}
-
}
-
}
-
else
-
{
-
for
(int
i
=
0
;
i
<
6
;
i++)
-
{
-
int
tx
=
now.x
+
dis[
i][
0]
;
-
int
ty
=
now.y
+
dis[
i][
1]
;
-
if
(tx
>=
1
&&
tx
<=
5
&&
ty
>=
1
&&
ty
<=
3
&&
g[
tx][
ty]
==
1
&&
book[
tx][
ty]
==
false
)
-
{
-
book[
tx][
ty]
=
true
;
-
q.push(Node(tx,
ty));
-
}
-
}
-
}
-
}
-
return
cnt
==
all;
-
}
-
void
draw(int
val)
-
{
-
if
(val
==
1
)
-
g[
1][
2]
=
1
;
-
if
(val
==
2
)
-
g[
2][
3]
=
1
;
-
if
(val
==
3
)
-
g[
4][
3]
=
1
;
-
if
(val
==
4
)
-
g[
5][
2]
=
1
;
-
if
(val
==
5
)
-
g[
4][
1]
=
1
;
-
if
(val
==
6
)
-
g[
2][
1]
=
1
;
-
if
(val
==
7
)
-
g[
3][
2]
=
1
;
-
return;
-
}
-
void
DFS(int
now,
int
sum)
-
{
-
if
(sum
==
C)
-
{
-
memset(g,
0,
sizeof
g);
-
for
(int
i
=
0
;
i
<
sum;
i++)
-
draw(a[
i]
);
-
if
(BFS(sum))
-
ans++;
-
return;
-
}
-
for
(int
i
=
now;
i
<=
7
;
i++)
-
{
-
if
(vis[
i]
==
false
)
-
{
-
vis[
i]
=
true
;
-
a[
sum]
=
i;
-
DFS(i
+
1,
sum
+
1
);
-
vis[
i]
=
false
;
-
}
-
}
-
return;
-
}
-
int
main()
-
{
-
-
for
(int
i
=
1
;
i
<=
7
;
i++)
-
{
-
C
=
i;
-
DFS(1,
0
);
-
}
-
cout
<<
ans;
-
return
0
;
-
}
80
F
-
// #pragma GCC optimize(2)
-
#include <iostream>
-
#include <cstdio>
-
#include <algorithm>
-
#include <queue>
-
#include <cmath>
-
#include <string>
-
#include <vector>
-
#include <stack>
-
#include <map>
-
#include <sstream>
-
#include <cstring>
-
#include <set>
-
#include <cctype>
-
#include <bitset>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
2e8 +
10;
-
const
int maxm =
2e5 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
6][
2] = {
1,
-1,
-1,
1,
-1,
-1,
1,
1,
2,
0,
-2,
0};
-
int m[
13] = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31};
-
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
IO;
-
int n, x;
-
int cnt1, cnt2 =
0;
-
char c =
'%';
-
cin >> n;
-
for (
int i =
0; i < n; i++)
-
{
-
cin >> x;
-
if (x >=
60)
-
cnt1++;
-
if (x >=
85)
-
cnt2++;
-
}
-
printf(
"%.0lf", round(
100.0 *
double(cnt1) /
double(n)));
-
printf(
"%c\n", c);
-
printf(
"%.0lf", round(
100.0 *
double(cnt2) /
double(n)));
-
printf(
"%c", c);
-
return
0;
-
}
G
当时抽疯写了个200+的弱智暴力（还不一定能对），其实只要枚举年份构造出回文串，然后把符合条件的日期存起来，二分答案即可。
H
太菜了只会暴力做，O（n^3)
是牛客多校上一道原题。
统计每个字符，对每个字符记录它上一次出现的位置vis，当前字符所能贡献的区间个数为（i-pre）*（n-i+1），i为当前字符下标。
-
// #pragma GCC optimize(2)
-
#include <algorithm>
-
#include <iostream>
-
#include <sstream>
-
#include <cstring>
-
#include <cstdio>
-
#include <random>
-
#include <cctype>
-
#include <bitset>
-
#include <string>
-
#include <vector>
-
#include <queue>
-
#include <cmath>
-
#include <stack>
-
#include <set>
-
#include <map>
-
#define IO \
-
ios::sync_with_stdio(false); \
-
// cout.tie(0);
-
using
namespace
std;
-
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
-
typedef
unsigned
long
long ULL;
-
typedef
long
long LL;
-
typedef
pair<
int,
int> P;
-
const
int maxn =
1e5 +
10;
-
const
int maxm =
1e6 +
10;
-
const LL INF =
0x3f3f3f3f3f3f3f3f;
-
const
int inf =
0x3f3f3f3f;
-
const LL mod =
1e9 +
7;
-
const
double eps =
1e-8;
-
const
double pi =
acos(
-1);
-
int dis[
4][
2] = {
1,
0,
0,
-1,
0,
1,
-1,
0};
-
// int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
-
LL vis[maxn];
-
char s[maxn];
-
int main()
-
{
-
#ifdef WXY
-
freopen(
"in.txt",
"r",
stdin);
-
// freopen("out.txt", "w", stdout);
-
#endif
-
scanf(
"%s", s +
1);
-
int n =
strlen(s +
1);
-
LL ans =
0;
-
for (
int i =
1; i <= n; i++)
-
{
-
int t = s[i] -
'a' +
1;
-
ans = ans + LL(i - vis[t]) * LL(n - i +
1);
-
vis[t] = i;
-
}
-
cout << ans;
-
return
0;
-
}
I
输出样例
当时如果仔细想想，应该能把n<=4的情况暴力模拟出来。。
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输出样例
不出意外，省一应该没了，还是太菜了。。。。
2020.10.27更新
居然省一了，本来以为填空全对才能摸到省一，没想到蓝桥杯这么水。。。
转载：https://blog.csdn.net/qq_44115065/article/details/109141107
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