Crazy mode is turned on the water solution to a problem, probably will last several times ... until I catch up with the progress so far.
The following explanations include:
\ [1001 【HDU-6624】 【1004 \\ HDU-6627】 【1005 \\ HDU-6628】 【1006 \\ HDU-6629】 【1007 \\ HDU-6630 [\]
[1001] math HDU-6624 fraction
http://acm.hdu.edu.cn/showproblem.php?pid=6624
Find the smallest positive integer \ (B \) satisfies \ (a <b \) and \ (A = BX (MOD \ P) \) .
Reference: https://blog.csdn.net/Sarah_Wang0220/article/details/98771865
Found: \ (0 <A <B \) , \ (A BX = Py + \) ==> \ (0 <= BX-Py A <B \) ==> \ (\ FRAC {X} {P} <\ FRAC {B} {Y} <\ FRAC {P}. 1-X} {\) , requiring minimal \ (B \) .
Iterative method:
\ [\ the aligned the begin {} & \ Because \ X FRAC {P} {} <\ FRAC {B} {Y} <\ FRAC {} {P}. 1-X \ and \ p> x \\ & \ therefore \ frac {p} { x}> 0 \ \ take \ t = \ frac {p} {x} \\ & \ therefore \ frac {p-tx} {x} <\ frac {b-ty} { y} <\ frac {pt ( x-1)} {x-1} \\ & obtained taking the inverse: \ frac {x-1} {pt (x-1)} <\ frac {y} {b-ty } <\ frac {x} { p-tx} \\ & Similarly, by subtracting the integer portion of the left continues \\ & case \ b '= b - ty decreasing \ end {aligned} \]
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
void solve(ll a, ll b, ll &c, ll &d, ll e, ll f) {
ll t = a/b;
if(e/f > t) {
c = t+1;
d = 1;
return ;
}
a = a - t*b;
e = e - t*f;
solve(f, e, d, c, b, a);
c = c + t*d;
}
int main() {
int t;
scanf("%d", &t);
while(t--) {
ll p, x;
scanf("%lld%lld", &p, &x);
ll b, y;
solve(p, x, b, y, p, x-1);
printf("%lld/%lld\n", b*x-p*y, b);
}
return 0;
}[1004] math HDU-6627 equation
http://acm.hdu.edu.cn/showproblem.php?pid=6627
Given \ (n-\) and \ (C \) , input \ (n-\) a \ (a_i \) and \ (B_i \) , is calculated for all \ (X \) Solutions such that: \ (\ SUM ^ { {I} _ = n-. 1} | + B_i a_ix | = C \) .
For each of the absolute value of the equation to find \ (x \) range, and then sort the enumeration range, the number of statistical solution.
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e5+5;
int n;
ll c;
ll suma[maxn], sumb[maxn]; // 记录 系数/常数 前缀和
map<double, int> vis;
struct node {
ll a, b;
double f;
bool operator < (const node &q) const {
return f > q.f;
}
}p[maxn], ans[maxn];
bool cmp(node x, node y) {
return x.f < y.f;
}
int main() {
int t;
scanf("%d", &t);
suma[0] = sumb[0] = 0;
while(t--) {
vis.clear();
scanf("%d%lld", &n, &c);
for(int i = 1; i <= n; i++) {
scanf("%lld%lld", &p[i].a, &p[i].b);
p[i].f = -1.0*p[i].b/p[i].a; // 每组方程等于 0的解
}
sort(p+1, p+1+n);
for(int i = 1; i <= n; i++) {
suma[i] = suma[i-1] + p[i].a;
sumb[i] = sumb[i-1] + p[i].b;
}
int flag = 0;
int cnt = 0;
for(int i = 0; i <= n; i++) {
ll A = suma[n] - 2*suma[i]; // 当前区间解的 系数 前缀和（分母）
ll B = sumb[n] - 2*sumb[i]; // 当前区间解的 常数 前缀和
ll C = c - B; // 等式常数移到右边获取最终常数（分子）
if(A == 0) {
if(C == 0) { // 无穷解
flag = 1;
break;
}
else { // 无解
continue;
}
}
else {
ll temp = __gcd(abs(C), abs(A));
double mark = 1.0 * C / A;
if(vis[mark] == 0) { // 去重
if((i==n || mark>p[i+1].f) && (i==0 || mark<=p[i].f)) { // 边界判定
ans[cnt].a = C / temp;
ans[cnt].b = A / temp;
ans[cnt++].f = mark;
}
}
}
}
if(flag == 1) {
printf("-1\n");
}
else {
printf("%d", cnt);
sort(ans, ans+cnt, cmp);
for(int i = 0; i < cnt; i++) {
if(ans[i].a*ans[i].b > 0) {
printf(" %lld/%lld", abs(ans[i].a), abs(ans[i].b));
}
else if(ans[i].a*ans[i].b == 0) {
printf(" 0/1");
}
else {
printf(" -%lld/%lld", abs(ans[i].a), abs(ans[i].b));
}
}
printf("\n");
}
}
return 0;
}[1005] thinking HDU-6628 permutation 1
http://acm.hdu.edu.cn/showproblem.php?pid=6628
\ (T \) of case, the known sequence is: \ (p_1p_2 ... P_n \) , defined as the difference in sequence: \ (P_1-P_2, P_3-P_2, ..., n--P_n-P_ {}. 1 \) . Given \ (n \) and \ (k \) , find the sequence \ (1,2,3, ..., n \ ) in small differences in the sequence that ranked K, output it. ( \ (2 \ n-Leq \ Leq 20 is \ \ \. 1 \ Leq K \ Leq min (10 ^. 4, n-!) \) )
For \ (n \ leq 8 \) , the direct violence on the line.
For \ (n> 8 \) time, find the law: to lay down \ (n \) this number, the latter is the smallest \ (, 2, 3, ..., the n--1 \) , and then only after 8 bits can be arranged, as \ (8!> 10 ^ 4 \) .
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 4e4+500;
int a[10] = {1, 2, 3, 4, 5, 6, 7, 8};
int a_8[10] = {1, 2, 3, 4, 5, 6, 7};
int _a_8[10] = {1, 2, 3, 4, 5, 6, 8};
int tot, tot1, tot2, cnt;
struct node {
int z[10];
bool operator < (const node x) const {
for(int i = 0; i < 8; i++) {
if(z[i] < x.z[i]) {
return 1;
}
else if(z[i] == x.z[i]) {
continue;
}
else {
return 0;
}
}
}
}temp[maxn], t_8[maxn], _t_8[maxn];
struct NODE {
int z[10];
int x[10];
}res[maxn];
bool cmp(NODE nn, NODE mm) {
for(int i = 0; i < 8; i++) {
if(nn.x[i] < mm.x[i]) {
return 1;
}
else if(nn.x[i] == mm.x[i]) {
continue;
}
else {
return 0;
}
}
return 0;
}
void init() {
do {
for(int i = 0; i < 8; i++) {
temp[tot].z[i] = a[i];
}
tot++;
} while(next_permutation(a, a+8));
sort(temp, temp+tot);
}
void cal(int n) {
int aa[10];
for(int i = 0; i < n; i++) {
aa[i] = i+1;
}
do {
for(int i = 0; i < n; i++) {
res[cnt].z[i] = aa[i];
}
for(int i = 1; i < n; i++) {
res[cnt].x[i-1] = res[cnt].z[i] - res[cnt].z[i-1];
// cout << res[cnt].x[i-1] << " ";
}
// cout << endl;
cnt++;
} while(next_permutation(aa, aa+n));
sort(res, res+cnt, cmp);
}
int main() {
tot = tot1 = tot2 = 0;
init();
int t;
scanf("%d", &t);
while(t--) {
int n, k;
scanf("%d%d", &n, &k);
if(n <= 8) {
cnt = 0;
cal(n);
for(int i = 0; i < n; i++) {
printf("%d%c", res[k-1].z[i], i==n-1?'\n':' ');
}
}
else {
printf("%d ", n);
for(int i = 1; i < n-8; i++) {
printf("%d ", i);
}
k = k - 1;
for(int i = 0; i < 8; i++) {
printf("%d%c", temp[k].z[i]+n-9, i==7?'\n':' ');
}
}
}
return 0;
}[1006] Extended KMP HDU-6629 string matching
http://acm.hdu.edu.cn/showproblem.php?pid=6629
Given a string, it is now required to have each of its common prefix and suffix original string, ask how much violence the implementation of the process steps required.
Extended KMP bare title.
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6+5;
char s[maxn];
char T[maxn];
ll nxt[maxn], extend[maxn];
void getnxt(char str[]) {
int i = 0, j, po;
int len = strlen(str);
nxt[0] = len;
while(str[i] == str[i+1] && i+1 < len) i++;
nxt[1] = i;
po = 1;
for(i = 2; i < len; i++) {
if(nxt[i-po]+i < nxt[po]+po) // case1 可以直接得到next[i]的值
nxt[i] = nxt[i-po];
else { // case2 要继续匹配才能得到next[i]的值
j = nxt[po] + po - i;
if(j < 0) // 如果i>po+next[po],则要从头开始匹配
j = 0;
while(str[j] == str[j+i] && i+j < len) j++;
nxt[i] = j;
po = i;
}
}
}
void EXKMP(char s1[], char s2[]) {
int i = 0, j, po;
int l1 = strlen(s1);
int l2 = strlen(s2);
getnxt(s2);
while(s1[i] == s2[i] && i < l2 && i < l1) i++;
extend[0] = i;
po = 0;
for(i = 1; i < l1; i++) {
if(nxt[i-po]+i < extend[po]+po) //case1 直接可以得到extend[i]的值
extend[i] = nxt[i-po];
else { // case2 要继续匹配才能得到extend[i]的值
j = extend[po]+po-i;
if(j < 0) // 如果i>extend[po]+po则要从头开始匹配
j = 0;
while(s1[j+i]==s2[j] && i+j < l1 && j < l2) j++;
extend[i] = j;
po = i;
}
}
}
int main() {
int t;
scanf("%d", &t);
while(t--) {
scanf("%s", s);
strcpy(T, s);
EXKMP(s, T);
int n = strlen(s);
ll ans = 0;
for(int i = 1; i < n; i++) {
ans = ans + extend[i];
if(extend[i] != n-i) {
ans = ans + 1;
}
}
printf("%lld\n", ans);
}
return 0;
}[1007] Law HDU-6630 permutation 2
http://acm.hdu.edu.cn/showproblem.php?pid=6630
Given \ (n-, X, Y \) , for \ (1 \) to the \ (n-\) This \ (n-\) number, satisfying: \ (P_1 = X, \ = Y P_2, \ | p_i- {I}. 1-P_ | \ Leq 2 \ (. 1 \ Leq I <n-) \) .
Hit the table to find the law can be.
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const int maxn = 1e5+5;
ll a[maxn];
int main() {
int t;
scanf("%d", &t);
while(t--) {
int n, x, y;
scanf("%d%d%d", &n, &x, &y);
if(n == 2 || n == 3) {
printf("1\n");
continue;
}
a[0] = 0;
if(x == 1) {
a[1] = 1;
a[2] = 1;
for(int i = 3; i <= y-x; i++) {
if(i == n-x) {
a[i] = (a[i-1] + a[i-2] + a[i-3]) % mod;
}
else {
a[i] = (a[i-1] + a[i-3]) % mod;
}
}
}
else if(x == n-1) {
printf("1\n");
continue;
}
else {
a[1] = 0;
a[2] = 1;
for(int i = 3; i <= y-x; i++) {
if(i == n-x) {
a[i] = (a[i-1] + a[i-2] + a[i-3]) % mod;
}
else {
a[i] = (a[i-1] + a[i-3]) % mod;
}
}
}
printf("%lld\n", a[y-x]);
}
return 0;
}