2019.10.18考试解题报告
2019.10.18考试解题报告
总结
期望得分：\(100 + 100 + 10\)
实际得分：\(0 + 20 + 10\)
完美.
今天的考试格外完美，\(T1\)做了*\(2.5h\)，最后换来了\(0\)分的好成绩，史无前例，美妙绝伦，我竟然不删调试，做得™好。
\(T2\)是个好题，是个好阅读题，\(n\)和\(m\)写反了，样例给的是\(n\)和\(m\)相等的情况，最终完美\(100—>20\)，我竟然这么粗心，题目竟然没读好，做得™好.
\(T3\)没时间了，都耗在\(T1\)上了，可惜\(T1\)还没有分，做得™好.
总体来说，™好，做得™好，做的太™好了
希望以后继续粗心马虎不看好程序，创下更好的成绩
（写下这段话，嘲讽我自己）
思路
T1
先写了\(30\)分暴力，直接\(O(n^4)\)枚举
然后写\(hash\)，调不出来，最后发现初始化\(hash\)的时候写错了，改完之后看着答案对，调试的东西也忘记删了，最终完美爆零
然而考完试之后，拿我的暴力程序去测试，却得了\(90\)分
意思就是……我在前半个小时，就打了\(90\)分！！
然后在最后\(10\)分上杠了两个小时！
最后还错了！
麻麻我要回家，社会太险恶了
一把辛酸泪啊！
T2
由题意可得，\(m\)是行数，\(n\)是列数，只走\(23\)个拐弯，\(n\)和\(m\)的范围又很小，所以就可以写\(n*m*23*4\)的\(dp\)了！
设\(f[i][j][k]\)表示在\((i, j)\)点走了\(k\)步获得的最大价值
因为每个点都可以重复进入，所以我们可以得到方程
\[f[i][j][k] = max(f[i][j][k], f[nx][ny][k - 1] + a[i][j])(nx,ny)是点(i, j)上下左右的四个点
\]
然后就做完了
T3
首先打表找规律:
1 0
2 1
3 2 = (1 + 0) * 2
4 9 = (1 + 2) * 3
5 44 = (9 + 2) * 4
6 265 = (44 + 9)* 5
7 1854 = (265 + 44) * 6
8 14833 = ...
9 133496
10 1334961
11 14684570
12 176214841
然后发现规律是\(f[n] = (f[n - 1] + f[n - 2]) *(i - 1)\)
最后套上高精就完了
代码
T1
考场爆零代码（可能会编译失败，这告诉我，变量用\(cnm\)也不用\(hash\)）
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define ull unsigned long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 211;
char pa[N][N], meng[N];
int n, m, a[N], l[N], ans, cnt[N];
ull p[N], hash[N][N], sjp[N];
//void work() {
// int len = strlen(meng + 1);
// for(int i = 1; i <= m; i++) {
// for(int j = 1; j <= len; j++) {
// if(meng[j] == pa[i][1]) {
// int sta = j, now = j, st = 1, flag = 0;
// while(meng[now] == pa[i][st] && st <= l[i] && now <= len) {
// now++, st++;
// flag = 1;
// }
// if(flag) now--, st--;
// if(st == l[i]) ans += sta * a[i];
// }
// }
// }
//}
bool query(int l1, int r1,int now) {
ull h1, h2;
h1 = sjp[r1] - sjp[l1 - 1] * p[r1 - l1 + 1];
h2 = hash[now][l[now]];
return h1 == h2;
}
void work2() {//hash
memset(sjp, 0, sizeof(sjp));
int len = strlen(meng + 1);
for(int i = 1; i <= len; i++) {
sjp[i] = sjp[i - 1] * 27 + meng[i] - 'a' + 1;
}
for(int i = 1; i <= len; i++) cout << sjp[i] << ' ';
cout << '\n';
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= len; j++) {
if(meng[j] == pa[i][1]) {
if(query(j, j + l[i] - 1, i)) ans += j * a[i];
}
}
}
}
int main() {
freopen("dream.in", "r", stdin);
freopen("dream.out", "w", stdout);
n = read(), m = read();
p[0] = 1;
for(int i = 1; i <= 200; i++) p[i] = p[i - 1] * 27ull;
for(int i = 1; i <= m; i++) scanf("%s", pa[i] + 1), l[i] = strlen(pa[i] + 1);
for(int i = 1; i <= m; i++)
for(int j = 1; j <= l[i]; j++)
hash[i][j] = hash[i][j - 1] * 27 + pa[i][j] - 'a' + 1;
for(int i = 1; i <= m; i++) a[i] = read();
for(int i = 1; i <= n; i++) scanf("%s", meng + 1), work2();
cout << ans << '\n';
return 0;
}
/*
1 1
a
1
aaaaaaa
2 2
abc
def
1
2
abcdef
defabc
*/
可以拿\(90\)的暴力
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define ull unsigned long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 211;
char pa[N][N], meng[N];
int n, m, a[N], l[N], ans, cnt[N];
ull p[N], cnm[N][N], sjp[N];
void work() {
int len = strlen(meng + 1);
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= len; j++) {
if(meng[j] == pa[i][1]) {
int sta = j, now = j, st = 1, flag = 0;
while(meng[now] == pa[i][st] && st <= l[i] && now <= len) {
now++, st++;
flag = 1;
}
if(flag) now--, st--;
if(st == l[i]) ans += sta * a[i];
}
}
}
}
int main() {
freopen("dream.in", "r", stdin);
freopen("dream.out", "w", stdout);
n = read(), m = read();
for(int i = 1; i <= m; i++) scanf("%s", pa[i] + 1), l[i] = strlen(pa[i] + 1);
for(int i = 1; i <= m; i++) a[i] = read();
for(int i = 1; i <= n; i++) scanf("%s", meng + 1), work();
cout << ans << '\n';
return 0;
}
/*
1 1
a
1
aaaaaaa
2 2
abc
def
1
2
abcdef
defabc
*/
\(hash\)满分
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define ull unsigned long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 211;
char pa[N][N], meng[N];
int n, m, a[N], l[N], ans, cnt[N];
ull p[N], cnm[N][N], sjp[N];
bool query(int l1, int r1,int now) {
ull h1, h2;
h1 = sjp[r1] - sjp[l1 - 1] * p[r1 - l1 + 1];
h2 = cnm[now][l[now]];
return h1 == h2;
}
void work2() {//cnm
memset(sjp, 0, sizeof(sjp));
int len = strlen(meng + 1);
for(int i = 1; i <= len; i++) {
sjp[i] = sjp[i - 1] * 27 + meng[i] - 'a' + 1;
}
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= len; j++) {
if(meng[j] == pa[i][1]) {
if(query(j, j + l[i] - 1, i)) ans += j * a[i];
}
}
}
}
int main() {
freopen("dream.in", "r", stdin);
freopen("dream.out", "w", stdout);
n = read(), m = read();
p[0] = 1;
for(int i = 1; i <= 200; i++) p[i] = p[i - 1] * 27ull;
for(int i = 1; i <= m; i++) scanf("%s", pa[i] + 1), l[i] = strlen(pa[i] + 1);
for(int i = 1; i <= m; i++)
for(int j = 1; j <= l[i]; j++)
cnm[i][j] = cnm[i][j - 1] * 27 + pa[i][j] - 'a' + 1;
for(int i = 1; i <= m; i++) a[i] = read();
for(int i = 1; i <= n; i++) scanf("%s", meng + 1), work2();
cout << ans << '\n';
return 0;
}
/*
1 1
a
1
aaaaaaa
2 2
abc
def
1
2
abcdef
defabc
*/
T2
考场\(20\)代码
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define int long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 311;
int n, m, sx, sy, a[N][N], f[N][N][23];
const int dx[4] = {0, 1, -1, 0};
const int dy[4] = {1, 0, 0, -1};
signed main() {
freopen("corner.in", "r", stdin);
freopen("corner.out", "w", stdout);
memset(f, -1, sizeof(f));
n = read(), m = read();
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
a[i][j] = read();
if(a[i][j] == 0) f[i][j][0] = 0;
}
}
for(int k = 1; k <= 23; k++) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
if(a[i][j] != -1 && a[i][j] != 0) {
for(int l = 0; l <= 3; l++) {
int nx = i + dx[l], ny = j + dy[l];
if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && a[nx][ny] >= 0 && f[nx][ny][k - 1] != -1) {
f[i][j][k] = max(f[i][j][k], f[nx][ny][k - 1] + a[i][j]);
}
}
}
}
}
}
int ans = -0x3f3f3f3f;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
ans = max(ans, f[i][j][23]);
}
}
cout << ans << "\n";
return 0;
}
改改\(m\)和\(n\)之后\(ac\)
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define int long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 311;
int n, m, sx, sy, a[N][N], f[N][N][23];
const int dx[4] = {0, 1, -1, 0};
const int dy[4] = {1, 0, 0, -1};
signed main() {
freopen("corner.in", "r", stdin);
freopen("corner.out", "w", stdout);
memset(f, -1, sizeof(f));
m = read(), n = read();
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
a[i][j] = read();
if(a[i][j] == 0) f[i][j][0] = 0;
}
}
for(int k = 1; k <= 23; k++) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
if(a[i][j] != -1 && a[i][j] != 0) {
for(int l = 0; l <= 3; l++) {
int nx = i + dx[l], ny = j + dy[l];
if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && a[nx][ny] >= 0 && f[nx][ny][k - 1] != -1) {
f[i][j][k] = max(f[i][j][k], f[nx][ny][k - 1] + a[i][j]);
}
}
}
}
}
}
int ans = -0x3f3f3f3f;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
ans = max(ans, f[i][j][23]);
}
}
cout << ans << "\n";
return 0;
}
T3
正解(\(from\ gyh dalao ? : zlq\))
#include<cstdio>
#include<iostream>
#include<vector>
#include<iomanip>
#include<cassert>
#include<algorithm>
#include<cstring>
#define rg register
#define LL long long
const int MARX = 1010;
const int Big_B = 10; const int Big_L = 1;
inline int intcmp_ (int a, int b) { if (a > b) return 1; return a < b ? -1 : 0; }
struct Int
{
inline int max(int a, int b) {return a > b ? a : b;}
inline int min(int a, int b) {return a < b ? a : b;}
std :: vector <int> c; Int () {}
Int (int x) { for (; x > 0; c.push_back (x % Big_B), x /= Big_B); }
Int (LL x) { for (; x > 0; c.push_back (x % Big_B), x /= Big_B); }
inline void CrZ () { for (; !c.empty () && c.back () == 0; c.pop_back ()); }
inline Int &operator += (const Int &rhs)
{
c.resize (max (c.size (), rhs.c.size ())); rg int i, t = 0, S;
for (i = 0, S = rhs.c.size (); i < S; ++ i)
c[i] += rhs.c[i] + t, t = c[i] >= Big_B, c[i] -= Big_B & (-t);
for (i = rhs.c.size (), S = c.size (); t && i < S; ++ i)
c[i] += t, t = c[i] >= Big_B, c[i] -= Big_B & (-t);
if (t) c.push_back (t); return *this;
}
inline Int &operator *= (const Int &rhs)
{
rg int na = c.size (), i, j, S, ai;
c.resize (na + rhs.c.size ()); LL t;
for (i = na - 1; i >= 0; -- i)
{
ai = c[i], t = 0, c[i] = 0;
for (j = 0, S = rhs.c.size (); j < S; ++ j)
{
t += c[i + j] + (LL) ai * rhs.c[j];
c[i + j] = t % Big_B, t /= Big_B;
}
for (j = rhs.c.size (), S = c.size (); t != 0 && i + j < S; ++ j)
t += c[i + j], c[i + j] = t % Big_B, t /= Big_B;
assert (t == 0);
}
CrZ (); return *this;
}
friend inline Int operator + (const Int &lhs, const Int &rhs)
{ Int res = lhs; return res += rhs; }
friend inline Int operator * (const Int &lhs, const Int &rhs)
{ Int res = lhs; return res *= rhs; }
friend inline std :: ostream &operator << (std :: ostream &out, const Int &rhs)
{
if (rhs.c.size () == 0) out << "0";
else
{
out << rhs.c.back ();
for (rg int i = rhs.c.size () - 2; i >= 0; -- i)
out << std :: setfill ('0') << std :: setw (Big_L) << rhs.c[i];
}
return out;
}
};
//=============================================================
LL n;
Int f[MARX];
//=============================================================
inline LL read()
{
LL s = 1, w=0; char ch=getchar();
for(; !isdigit(ch); ch=getchar()) if(ch=='-') s =-1;
for(; isdigit(ch); ch=getchar()) w = w*10+ch-'0';
return s*w;
}
//=============================================================
signed main()
{
freopen("keke.in","r",stdin);
freopen("keke.out","w",stdout);
n = read();
f[1] = 0, f[2] = 1;
if(n <= 2) { std::cout << f[n]; return 0; }
for(rg int i = 3; i <= n; i ++) f[i] = (f[i - 1] + f[i - 2]) * (i - 1);
std::cout << f[n];
}
我这个菜鸡的代码（高精我不会）
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 3010;
struct bign {
int d[N], len;
void clean() { while (len > 1 && !d[len - 1]) len--; }
bign() { memset(d, 0, sizeof(d)); len = 1; }
bign(int num) { *this = num; }
bign(char* num) { *this = num; }
bign operator = (const char* num) {
memset(d, 0, sizeof(d)); len = strlen(num);
for (int i = 0; i < len; i++) d[i] = num[len - 1 - i] - '0';
clean();
return *this;
}
bign operator = (int num) {
char s[20]; sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator + (const bign& b) {
bign c = *this; int i;
for (i = 0; i < b.len; i++) {
c.d[i] += b.d[i];
if (c.d[i] > 9) c.d[i] %= 10, c.d[i + 1]++;
}
while (c.d[i] > 9) c.d[i++] %= 10, c.d[i]++;
c.len = max(len, b.len);
if (c.d[i] && c.len <= i) c.len = i + 1;
return c;
}
bign operator * (const bign& b)const {
int i, j; bign c; c.len = len + b.len;
for (j = 0; j < b.len; j++) for (i = 0; i < len; i++)
c.d[i + j] += d[i] * b.d[j];
for (i = 0; i < c.len - 1; i++)
c.d[i + 1] += c.d[i] / 10, c.d[i] %= 10;
c.clean();
return c;
}
string str() const {
char s[N] = {};
for (int i = 0; i < len; i++) s[len - 1 - i] = d[i] + '0';
return s;
}
};
ostream& operator << (ostream& out, const bign& x) {
out << x.str();
return out;
}
int n;
bign ans, f[N];
signed main() {
freopen("keke.in", "r", stdin);
freopen("keke.out", "w", stdout);
cin >> n;
f[1] = 0, f[2] = 1;
if(n <= 2) return cout << f[n] << '\n', 0;
for (int i = 3; i <= n; ++i) {
f[i] = (f[i - 1] + f[i - 2]) * (i - 1);
}
cout << f[n] << endl;
return 0;
}
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