K小割
题目链接：WC2015 K小割
Description
题目很清楚了，已经不能说的更简洁了……
Solution
这道题出题人挺毒的，你需要针对不同的部分分施用不同的做法 。
- 第(1)部分：暴力枚举每条边是否割掉，并保留所有合法的割，更新答案，最后(sort)一下（从小到大），输出即可。
复杂度(O(2^m imes n))，可以规避(K)过大的限制。 - 第(2)部分：采用优先队列。
对于除(s,t)外的每个点都分别有且仅有一条边与(s,t)相连。那么对于每一个点都必须至少割掉其中的一条边。设两条边的权值为(a,b (a<b))，那么对于每个点的选择有({a,b,a+b}) 三种，我们可以通过对于每个点选择的升级/降级，来达到不同的状态。
我们将点以((b-a))为第一关键字（从第一级升到第二级的代价），(a)为第二关键字排序（从第二级升到第三级的代价）。
对于每个状态我们有三种选择
（1）将当前点的状态升级
（2）当前点保持不变，将该点的后一个点升级，并将当前点改为他后面的点。
（3）将当前点降级，将该点的后一个点升级，并将当前点改为他后面的点（需要注意的是在降级的时候只能是从(b->a)，不能是从(a+b->b)，否则会出现重复的状态，因为(a+b->b)的改变其实可以转化成选择（2））
总之在更新状态的时候需要时刻注意，不要更新到之前已经得到的状态，如果开始后一个点的升级，那么当前点就不能再改变了。 - 第(3)部分：采用最小割。
我们可以先跑最大流，求出最小割。
然后，我们可以按如下方式求出次小割：
((1)) 强制割集中的某条边不选，然后求此时的最小割；
((2)) 选择不再割集中的某条边，然后割掉。
对于第（2）种产生方式，必然是选择不在割集中的边权最小的边来更新答案。那么我们如何确定割集呢？从s开始走剩余流量不为0的边，将所有能遍历到的点打访问标记，如果一条边的两个端点x[i]有标记，y[i]没有标记，那么该边在最小割的割集中。
第一种情况，对于每一条边来说，如果这条边不割，那么(s->x[i]，y[i]->t)都其中一个必须不连通，我们可以在做完最小割的残量网络上对于(s->x[i],y[i]->t)分别求最小割，然后从中选取较小的更新答案。
处理完((1)(2))两种产生方式后，得到的答案就是针对当前割集的次小割。
结合这(3)部分，你就可以获得(AC)。
混淆与破解
题目链接：WC2015 混淆与破解
Description
Solution
Code
这道题还不会，暂时咕着。
未来程序
题目链接：
Description
这是一道提交答案题
题目给定你(10)组(program*.cpp)和(program*.in)，这些都是暴力程序，无法在规定时间内跑出。
你需要优化这(10)个程序，并提交对应的(10)组(program*.out)。
Solution
- program1
#include <stdio.h>
#include <stdlib.h>
void _() {
unsigned long long a, b, c, d, i;
scanf("%llu %llu %llu", &a, &b, &c);
i = 0;
d = 0;
while (i < b) {
d = d + a;
d = d % c;
i = i + 1;
}
printf("%llu
", d);
}
int main() {
_();
_();
_();
_();
_();
_();
_();
_();
_();
_();
}
这个很显然是求(a imes b mod c)，我们用快速乘优化即可。
(program1.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
unsigned long long a, b, c;
ull fmul(ull a, ull b, ull c) {
ull ans = 0;
while (b > 0) {
if (b & 1) ans = (ans + a) % c;
a = (a + a) % c;
b >>= 1;
}
return ans;
}
int main() {
freopen("program1.in", "r", stdin);
freopen("program1.out", "w", stdout);
while (cin >> a >> b >> c) {
cout << fmul(a, b, c) << '
';
}
return 0;
}
- program2
#include <stdio.h>
void _______() {
long long i, n, a, b, c, d, p;
a = 1;
b = 0;
c = 0;
scanf("%lld %lld", &n, &p);
i = 1;
while (i <= n) {
i = i + 1;
b = a + b;
a = 2 * b - a + c;
c = 2 * b - a + c;
while (a >= p) {
a -= p;
}
while (a < 0) {
a += p;
}
while (b >= p) {
b -= p;
}
while (b < 0) {
b += p;
}
while (c >= p) {
c -= p;
}
while (c < 0) {
c += p;
}
}
d = a - 2 * b + c;
while (d >= p) {
d -= p;
}
while (d < 0) {
d += p;
}
printf("%lld
", d);
}
int main() {
_______();
_______();
_______();
_______();
_______();
_______();
_______();
_______();
_______();
_______();
return 0;
}
我们发现这其实就是求以(0 1)开头的斐波那契数列第(n)项的平方。
矩阵快速幂优化即可。
(program2.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
long long n, mod;
struct Matrix {
ll v[2][2];
int n, m;
Matrix (int _n = 0, int _m = 0) {
n = _n;
m = _m;
}
void clear() {
memset(v, 0, sizeof(v));
}
} ans(2, 1), res(2, 2);
Matrix operator * (Matrix a, Matrix b) {
Matrix ans(a.n, b.m);
ans.clear();
for (rint i = 0; i < a.n; i++) {
for (rint j = 0; j < b.m; j++) {
for (rint k = 0; k < a.m; k++) {
(ans.v[i][j] += a.v[i][k] * b.v[k][j]) %= mod;
}
}
}
return ans;
}
ll fib(ll n) {
if (n == 1) return 0;
if (n == 2) return 1;
ans.clear(), res.clear();
ans.v[0][0] = 1, ans.v[1][0] = 0;
res.v[0][0] = res.v[0][1] = res.v[1][0] = res.v[0][0] = 1;
n -= 2;
while (n > 0) {
if (n & 1) ans = res * ans;
res = res * res;
n >>= 1;
}
return ans.v[0][0];
}
int main() {
freopen("program2.in", "r", stdin);
freopen("program2.out", "w", stdout);
while (cin >> n >> mod) {
ll x = fib(n);
cout << x * x % mod << '
';
}
return 0;
}
- program3
#include <stdio.h>
unsigned long long s0, s1, s2, s3, s4, i, n;
int main() {
scanf("%llu", &n);
i = 0;
while (i <= n) {
s0 = s0 + 1;
s1 = s1 + i;
s2 = s2 + i * i;
s3 = s3 + i * i * i;
s4 = s4 + i * i * i * i;
i = i + 1;
}
printf("%llu
", s0);
printf("%llu
", s0);
printf("%llu
", s1);
printf("%llu
", s1);
printf("%llu
", s2);
printf("%llu
", s2);
printf("%llu
", s3);
printf("%llu
", s3);
printf("%llu
", s4);
printf("%llu
", s4);
return 0;
}
这是让我们求
[s0=sum_{i=1}^{n}1
]
[s1=sum_{i=1}^{n}i
]
[s2=sum_{i=1}^{n}i^2
]
[s3=sum_{i=1}^{n}i^3
]
[s4=sum_{i=1}^{n}i^4
]
直接套数学公式即可，四次方和公式可别忘了呐~
(program3.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
ull n;
int main() {
freopen("program3.in", "r", stdin);
freopen("program3.out", "w", stdout);
while (cin >> n) {
ull ans = n + 1;
cout << ans << '
';
cout << ans << '
';
ans = n * (n + 1) / 2ull;
cout << ans << '
';
cout << ans << '
';
ans = n * (n + 1) * (2 * n + 1) / 6ull;
cout << ans << '
';
cout << ans << '
';
ans = n * (n + 1) / 2ull;
ans = ans * ans;
cout << ans << '
';
cout << ans << '
';
ull a[5] = {n, n + 1, 2 * n + 1, 3 * n * n + 3 * n - 1};
ull ne[5] = {2, 3, 5};
for (rint i = 0; i < 3; i++) {
for (rint j = 0; j < 4; j++) {
if (a[j] % ne[i] == 0) {
a[j] /= ne[i];
break;
}
}
}
ans = a[0] * a[1] * a[2] * a[3];
//ans = n * (n + 1) * (2ull * n + 1ull) / 30ull * (3ull * n * n + 3ull * n - 1ull);
cout << ans << '
';
cout << ans << '
';
}
}
- program4
#include <iostream>
const int N = 5000, inf = 0x3F3F3F3F;
int n, m, type;
bool data[N + 11][N + 11];
int seed;
int next_rand(){
static const int P = 1000000007, Q = 83978833, R = 8523467;
return seed = ((long long)Q * seed % P * seed + R) % P;
}
void generate_input(){
std::cin >> n >> m >> type;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
data[i][j] = bool((next_rand() % 8) > 0);
}
long long count1(){
long long ans = 0LL;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(data[i][j])
for(int k = 0; k < n; k++)
for(int l = 0; l < m; l++)
if(data[k][l] && (k != i || l != j))
ans++;
return ans;
}
int abs_int(int x){
return x < 0 ? -x : x;
}
long long count2(){
long long ans = 0LL;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(data[i][j]){
int level = inf;
for(int k = 0; k < n; k++)
for(int l = 0; l < m; l++)
if(!data[k][l]){
int dist = abs_int(k - i) + abs_int(l - j);
if(level > dist)
level = dist;
}
ans += level;
}
return ans;
}
int main(){
std::cin >> seed;
for(int i = 0; i < 10; i++){
generate_input();
std::cout << (type == 1 ? count2() : count1()) << std::endl;
}
return 0;
}
这道题如果(type=0)，那么就是求有多少对二元组(((x_i,y_i),(x_j,y_j)) (i≠j))，满足(data_{x_i,y_i}=1)且(data_{x_j,y_j}=1)。
如果(type=1)，那么就是求每一个(x_i,y_i)满足(data_{x_i,y_i}=1)，离它曼哈顿距离最近的(data_{x_j,y_j}=0)距离之和。
显然直接(dp)即可。
(program4.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int N = 5005;
int data[N][N];
int seed, n, m, type;
int next_rand() {
static const int P = 1000000007, Q = 83978833, R = 8523467;
return seed = ((long long)Q * seed % P * seed + R) % P;
}
void generate_input() {
cin >> n >> m >> type;
for (rint i = 1; i <= n; i++) {
for (rint j = 1; j <= m; j++) {
data[i][j] = bool((next_rand() % 8) > 0);
}
}
}
long long count1() {
long long ans = 0ll;
for (rint i = 1; i <= n; i++) {
for (rint j = 1; j <= m; j++) {
if (data[i][j] == 1) {
ans++;
}
}
}
return ans * (ans - 1);
}
int dp1[N][N];
int dp2[N][N];
int dp3[N][N];
int dp4[N][N];
long long count2() {
mset(dp1, 0x3f), mset(dp2, 0x3f), mset(dp3, 0x3f), mset(dp4, 0x3f);
for (rint i = 1; i <= n; i++) {
for (rint j = 1; j <= m; j++) {
if (!data[i][j]) dp1[i][j] = 0;
else dp1[i][j] = min(dp1[i - 1][j], dp1[i][j - 1]) + 1;
}
}
for (rint i = 1; i <= n; i++) {
for (rint j = m; j >= 1; j--) {
if (!data[i][j]) dp2[i][j] = 0;
else dp2[i][j] = min(dp2[i - 1][j], dp2[i][j + 1]) + 1;
}
}
for (rint i = n; i >= 1; i--) {
for (rint j = 1; j <= m; j++) {
if (!data[i][j]) dp3[i][j] = 0;
else dp3[i][j] = min(dp3[i + 1][j], dp3[i][j - 1]) + 1;
}
}
for (rint i = n; i >= 1; i--) {
for (rint j = m; j >= 1; j--) {
if (!data[i][j]) dp4[i][j] = 0;
else dp4[i][j] = min(dp4[i + 1][j], dp4[i][j + 1]) + 1;
}
}
long long ans = 0ll;
for (rint i = 1; i <= n; i++) {
for (rint j = 1; j <= m; j++) {
if (data[i][j]) {
ans += min(min(dp1[i][j], dp2[i][j]), min(dp3[i][j], dp4[i][j]));
}
}
}
return ans;
}
int main() {
freopen("program4.in", "r", stdin);
freopen("program4.out", "w", stdout);
cin >> seed;
for (rint i = 0; i < 10; i++) {
generate_input();
cout << (type == 1 ? count2() : count1()) << endl;
}
return 0;
}
- program5
#include <iostream>
const int N = 5011;
int n, m;
bool data[N][N];
int seed;
int next_rand(){
static const int P = 1000000007, Q = 83978833, R = 8523467;
return seed = ((long long)Q * seed % P * seed + R) % P;
}
void generate_input(){
std::cin >> n >> m;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
data[i][j] = bool((next_rand() % 8) > 0);
}
bool check(int x1, int y1, int x2, int y2){
bool flag = true;
for(int i = x1; i <= x2; i++)
for(int j = y1; j <= y2; j++)
if(!data[i][j])
flag = false;
return flag;
}
long long count3(){
long long ans = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
for(int k = i; k < n; k++)
for(int l = j; l < m; l++)
if(check(i, j, k, l))
ans++;
return ans;
}
int main(){
std::cin >> seed;
for(int i = 0; i < 10; i++){
generate_input();
std::cout << count3() << std::endl;
}
return 0;
}
题目问你有多少个内部全是(1)的矩阵。
我们先表示：
(h_{i,j})表示((i,j))这个点往上连续(1)的最长长度。
(l_{i,j})表示从((i,j))开始第一个满足(h_{i,l_{i,j}}le h(i,j))的点，如果不存在，令(l_{i,j}=0)。
(r_{i,j})表示从((i,j))开始第一个满足(h_{i,r_{i,j}}le h(i,j))的点，如果不存在，令(r_{i,j}=m+1)。
那么，对于((i,j))为矩形底的贡献，就是(val=(j-l_{i,j})*(r_{i,j}-j)*h_{i,j})。
考虑一下，这样做如何保证答案不重不漏。
不重：当且仅当在同一行存在两个数(l_{i,j_1}=l_{i,j_2})并且(r_{i,j_1}=r_{i,j_2})的时候，才有可能算重矩形。
但是这种情况是不存在的，因为(l_{i,j})满足了左边第一个小于等于的，右边第一个小于的，显然无法构造出这种情况。
不漏：对于一个矩形，总有一个(l_{i,j},r_{i,j})能框住一个矩形的两边，故这个矩形一定能被计算到。
我们可以通过一个单调栈来计算(l_{i,j})和(r_{i,j})，复杂度(O(n^2))。
但是我写代码的时候(sb)了，所以用了一个单调队列来维护，但本质上是一样的。
(program5.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int N = 5005;
int a[N][N], h[N][N], l[N][N], r[N][N];
int seed, n, m;
int next_rand() {
static const int P = 1000000007, Q = 83978833, R = 8523467;
return seed = ((long long)Q * seed % P * seed + R) % P;
}
void generate_input() {
cin >> n >> m;
for (rint i = 1; i <= n; i++) {
for (rint j = 1; j <= m; j++) {
a[i][j] = bool((next_rand() % 8) > 0);
}
}
}
deque <int> deq;
void push_l(int i, int j) {
while (!deq.empty() && h[i][deq.back()] > h[i][j]) r[i][deq.back()] = j, deq.pop_back();
deq.push_back(j);
}
void push_r(int i, int j) {
while (!deq.empty() && h[i][deq.back()] >= h[i][j]) l[i][deq.back()] = j, deq.pop_back();
deq.push_back(j);
}
long long count3() {
for (rint j = 1; j <= m; j++) {
for (rint i = 1; i <= n; i++) {
if (a[i][j]) h[i][j] = h[i - 1][j] + 1;
else h[i][j] = 0;
}
}
long long ans = 0ll;
for (rint i = 1; i <= n; i++) {
while (!deq.empty()) deq.pop_back();
for (rint j = 1; j <= m; j++) {
push_l(i, j);
}
while (!deq.empty()) r[i][deq.back()] = m + 1, deq.pop_back();
for (rint j = m; j >= 1; j--) {
push_r(i, j);
}
while (!deq.empty()) l[i][deq.back()] = 0, deq.pop_back();
for (rint j = 1; j <= m; j++) {
//printf("l[%d][%d] = %d, r[%d][%d] = %d
", i, j, l[i][j], i, j, r[i][j]);
//printf("h[%d][%d] = %d
", i, j, h[i][j]);
ans += 1ll * (j - l[i][j]) * (r[i][j] - j) * h[i][j];
}
}
return ans;
}
int main() {
freopen("program5.in", "r", stdin);
freopen("program5.out", "w", stdout);
cin >> seed;
for (rint i = 0; i < 10; i++) {
generate_input();
cout << count3() << endl;
}
return 0;
}
- program6
#include <stdio.h>
unsigned long long a, b, c, t, k, n;
unsigned long long rd() {
t = (t * t * a + b) % c;
return t;
}
int main() {
int i;
for (i = 0; i <= 9; i++) {
scanf("%llu %llu %llu %llu", &n, &a, &b, &c);
t = 0;
k = 1;
while (k <= n) {
k = k + 1;
rd();
}
printf("%llu
", t);
}
}
这道题让你计算(n)次(t=(t*t*a+b)\%c)，很显然直接做是不行的。
我们考虑用(Floyd)判圈法。
形象的说，就是乌龟赛跑。乌龟每次走(1)步，兔子每次跑(2)步，如果这个(t)的取值存在循环，即图存在环，那么乌龟和兔子就会相遇。
具体见这篇博客，有讲两种判圈算法。
(program6.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
ull n, a, b, c;
void nxt(ull &t) {
t = (t * t * a + b) % c;
}
int main() {
freopen("program6.in", "r", stdin);
freopen("program6.out", "w", stdout);
for (rint t = 0; t <= 9; t++) {
scanf("%llu%llu%llu%llu", &n, &a, &b, &c);
int tot = 0;
ull x = 0, y = 0;
do {
tot++;
nxt(x), nxt(y), nxt(y);
} while (x != y);
cerr << "meet
";
int cycle = 0;
do {
cycle++;
nxt(x);
} while (x != y);
cerr << "cycle = " << cycle << '
';
int turns = (n - tot) % cycle;
while (turns--) {
nxt(y);
}
printf("%llu
", y);
}
return 0;
}
- program7
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
char s[20][20];
bool check() {
for (int i = 0; i <= 15; i++) {
bool v[16];
memset(v, false, sizeof(v));
for (int j = 0; j <= 15; j++) {
v[s[i][j] - 'A'] = true;
}
for (int j = 0; j <= 15; j++) {
if (!v[j]) {
return false;
}
}
}
for (int i = 0; i <= 15; i++) {
bool v[16];
memset(v, false, sizeof(v));
for (int j = 0; j <= 15; j++) {
v[s[j][i] - 'A'] = true;
}
for (int j = 0; j <= 15; j++) {
if (!v[j]) {
return false;
}
}
}
for (int i = 0; i <= 15; i++) {
bool v[16];
memset(v, false, sizeof(v));
for (int j = 0; j <= 15; j++) {
v[s[i / 4 * 4 + j / 4][i % 4 * 4 + j % 4] - 'A'] = true;
}
for (int j = 0; j <= 15; j++) {
if (!v[j]) {
return false;
}
}
}
return true;
}
bool dfs(int x, int y) {
if (x == 16 && y == 0) {
return check();
}
if (s[x][y] == '?') {
for (char i = 'A'; i <= 'P'; i++) {
s[x][y] = i;
if (dfs(x, y)) {
return true;
}
s[x][y] = '?';
}
return false;
} else {
return dfs(x + (y + 1) / 16, (y + 1) % 16);
}
}
void solve(int points) {
for (int i = 0; i <= 15; i++) {
scanf("%s", s[i]);
}
if (dfs(0, 0)) {
for (int k = 0; k <= points - 1; k++) {
for (int i = 0; i <= 15; i++) {
printf("%s", s[i]);
}
putchar('
');
}
} else {
for (int k = 0; k <= points - 1; k++) {
printf("NO SOLUTION.
");
}
}
}
int main() {
solve(1);
solve(2);
solve(3);
solve(4);
return 0;
}
这是……“字母”独？
给了你一个(16 imes 16)的矩阵，你需要填上(A-P)，使得每行每列以及16个(4 imes 4)的宫内刚好是(A-P)这16个字母。
正着扫当然(TLE)，于是我们考虑倒着扫~
倒着扫竟然只要(3-)秒，太神仙了！
PS: 看来以后爆搜的题目倒着扫优秀点 /cy
(program7.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int N = 50;
char s[N][N];
int pw[N], row[N], col[N], block[N];
int be(int x, int y) {
return 4 * (x / 4) + (y / 4);
}
int opt = 0;
void dfs(int x, int y) {
if (x == -1) {
opt = 1;
return ;
}
if (opt) return ;
if (!opt && s[x][y] != '?') {
if (y == 0) dfs(x - 1, 15);
else dfs(x, y - 1);
} else {
for (rint i = 0; i <= 15; i++) {
if (opt) return ;
if (!opt && !((row[x] >> i) & 1) && !((col[y] >> i) & 1) && !((block[be(x, y)] >> i) & 1)) {
row[x] ^= pw[i];
col[y] ^= pw[i];
block[be(x, y)] ^= pw[i];
s[x][y] = 'A' + i;
if (y == 0) dfs(x - 1, 15);
else dfs(x, y - 1);
if (opt) return ;
s[x][y] = '?';
row[x] ^= pw[i];
col[y] ^= pw[i];
block[be(x, y)] ^= pw[i];
}
}
}
}
void solve(int points) {
memset(row, 0, sizeof(row));
memset(col, 0, sizeof(col));
memset(block, 0, sizeof(block));
for (rint i = 0; i <= 15; i++) {
scanf("%s", s[i]);
for (rint j = 0; j <= 15; j++) {
if (s[i][j] != '?') {
row[i] ^= pw[s[i][j] - 'A'];
col[j] ^= pw[s[i][j] - 'A'];
block[be(i, j)] ^= pw[s[i][j] - 'A'];
}
}
}
opt = 0, dfs(15, 15);
if (opt == 1) {
for (rint k = 0; k <= points - 1; k++) {
for (rint i = 0; i <= 15; i++) {
printf("%s", s[i]);
}
putchar('
');
}
} else {
for (rint k = 0; k <= points - 1; k++) {
puts("NO SOLUTION.");
}
}
}
int main() {
freopen("program7.in", "r", stdin);
freopen("program7.out", "w", stdout);
pw[0] = 1;
for (rint i = 1; i <= 15; i++) pw[i] = pw[i - 1] << 1;
solve(1), solve(2), solve(3), solve(4);
return 0;
}
- program8
#include <stdio.h>
unsigned long long a, b, c, d, e, f, g, n, q, r, s, t, u, v, w, x, y, z;
unsigned long long p = 1234567891;
int main() {
scanf("%llu", &n);
a = 0;
q = 0;
r = 0;
s = 0;
t = 0;
u = 0;
v = 0;
w = 0;
x = 0;
y = 0;
z = 0;
a = 0;
while (a < n) {
a = a + 1;
b = 0;
while (b < n) {
b = b + 1;
c = 0;
while (c < n) {
c = c + 1;
d = 0;
while (d < n) {
d = d + 1;
e = 0;
while (e < n) {
e = e + 1;
f = 0;
while (f < n) {
f = f + 1;
g = 0;
while (g < n) {
g = g + 1;
if (a < b && b < c && c < d && d < e && e < f && f < g) {
q = q + 1;
q = q % p;
}
if (a < b && c < g && c < d && e < f && a < d) {
r = r + 1;
r = r % p;
}
if (a < d && d < f && c < f && c < e && b < d) {
s = s + 1;
s = s % p;
}
if (d < e && b < d && a < f && d < e && b < g) {
t = t + 1;
t = t % p;
}
if (c < f && b < f && b < c && f < g && b < f) {
u = u + 1;
u = u % p;
}
if (b < d && b < c && d < f && c < e && b < e) {
v = v + 1;
v = v % p;
}
if (a < c && a < b && c < e && b < f && e < g) {
w = w + 1;
w = w % p;
}
if (b < d && b < f && a < g && c < g && a < e) {
x = x + 1;
x = x % p;
}
if (b < f && a < c && c < d && a < c && b < e) {
y = y + 1;
y = y % p;
}
if (d < e && e < f && a < d && c < g && b < d) {
z = z + 1;
z = z % p;
}
}
}
}
}
}
}
}
printf("%llu
", q);
printf("%llu
", r);
printf("%llu
", s);
printf("%llu
", t);
printf("%llu
", u);
printf("%llu
", v);
printf("%llu
", w);
printf("%llu
", x);
printf("%llu
", y);
printf("%llu
", z);
return 0;
}
仔细观察，我们就会发现，其实程序就是求了10组组合问题。组合数随便搞搞就可以得到答案。
(program8.cpp)
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
ull n, p = 1234567891;
ull ksm(ull a, ull b) {
ull res = 1;
while (b > 0) {
if (b & 1) res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
int main() {
freopen("program8.in", "r", stdin);
freopen("program8.out", "w", stdout);
scanf("%llu", &n);
n %= p;
printf("%I64u
",n>6?(n*(n-1)%p*(n-2)%p*(n-3)%p*(n-4)%p*(n-5)%p*(n-6)%p*ksm(5040,p-2)%p):0);
printf("%I64u
",n>1?((n-1)*(n-1)%p*n%p*n%p*(2*n-1)%p*(2*n%p*n%p-2*n%p+1+p)%p*ksm(60,p-2)%p):0);
printf("%I64u
",n>2?(n*n%p*(n-1)%p*(n-1)%p*(n-2)%p*(2*n-1)%p*(7*n-3)%p*ksm(360,p-2)%p):0);
printf("%I64u
",n>2?(n*n%p*n%p*(n-1)%p*(n-1)%p*(n-2)%p*(3*n-1)%p*ksm(48,p-2)%p):0);
printf("%I64u
",n>3?(n*n%p*n%p*n%p*(n-1)%p*(n-2)%p*(n-3)%p*ksm(24,p-2)%p):0);
printf("%I64u
",n>2?(n*n%p*n%p*(n-1)%p*(n-2)%p*(3*n*n%p-6*n%p+1+p)%p*ksm(60,p-2)%p):0);
printf("%I64u
",n>3?(n*n%p*(n-1)%p*(n-2)%p*(n-3)%p*(5*n*n%p-9*n%p+1+p)%p*ksm(360,p-2)%p):0);
printf("%I64u
",n>1?(n*n%p*(n-1)%p*(n-1)%p*(2*n-1)%p*(5*n%p*n%p-5*n%p+2+p)%p*ksm(144,p-2)%p):0);
printf("%I64u
",n>2?(n*n%p*n%p*(n-1)%p*(n-1)%p*(n-2)%p*(2*n-1)%p*ksm(36,p-2)%p):0);
printf("%I64u
",n>3?(n*n%p*(n-1)%p*(n-1)%p*(n-2)%p*(n-3)%p*(2*n-3)%p*ksm(240,p-2)%p):0);
return 0;
}
最后两组是恶搞（(program9)扯出(chenlijie)大神和(MD5)解密，(program10)扯出《独立宣言》，数每个字母出现的次数，并进行(hash)）