模拟赛小结：2017 China Collegiate Programming Contest Final (CCPC-Final 2017)
比赛链接：传送门
前期大顺风，2：30金区中游。后期开题乏力，掉到银尾。4：59绝杀I，但罚时太高卡在银首。
Problem A - Dogs and Cages 00：09：45 （+） Solved by Dancepted
算了半天发现就是n-1，被队友喷死，差点气哭。
Problem E - Evil Forest 00：16：54 （+） Solved by xk
xk大喊签到，就过了。
代码：
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } int main() {
fast;
int T, kase = ;
cin >> T;
while (T--) {
int n;
cin >> n;
int ans = ;
for(int i = ; i < n; i++)
{
int x;
cin >> x;
ans += x + upperdiv(x, );
}
cout << "Case #" << (kase++) << ": " << ans << endl;
}
return ;
}
Problem C - Rich Game 00：32：42 （+） Solved by xk
xk又喊了一声签到！
代码：
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } int main() {
fast;
int T, kase = ;
cin >> T;
while (T--) {
int x, y, k;
cin >> x >> y >> k;
int money = , ans = ;
if(x > y) ans = k;
else
{
for(int i = ; i < k; i++)
{
if(money + * x >= * y) {
money += * x - * y;
ans++;
}
else {
money += * x;
}
}
}
cout << "Case #" << (kase++) << ": " << ans << endl;
}
return ;
}
Problem K - Knightmare 00：46：33 （+） Solved by Dancepted
步数较少的时候可以往回走，把没走过的地方填上，步数增长到一定程度时，答案差不多会均匀增长。
我猜测是小数据打表，大数据是个公式之类的。
开完A趁电脑没人就打了个表，发现和我的猜测是一致的，果断交了一发就过了。
代码：
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <stdio.h>
#include <utility>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 1005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef __int128 ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } const int ans[] = {, , , , , };
int main() {
int T; cin >> T;
int kase = ;
while (T--) {
ll n; read(n);
printf("Case #%d: ", kase++);
if (n <= ) {
write(ans[n]);
}
else {
ll res = * n * n - * n + ;
write(res);
}
putchar('\n');
} return ;
}
Problem J - Subway Chasing 01：42：00 （+） Solved by xk & lh
好像是差分约束什么的。图论这种事相信队友就完了。
代码：
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } const int maxn = + ; struct edge
{
int v, l;
}; ll dis[maxn];
int cnt[maxn];
bool inq[maxn];
vector<edge> g[maxn]; void addedge(int u, int v, int l)
{
g[u].push_back(edge{v, l});
// g[v].push_back(edge{u, l});
} int n, m, x; bool spfa()
{
queue<int> q;
dis[] = ;
q.push();
cnt[] = inq[] = ;
while(!q.empty())
{
int u = q.front();
inq[u] = ;
q.pop();
for(auto &e : g[u])
{
if(dis[e.v] > dis[u] + e.l)
{
dis[e.v] = dis[u] + e.l; if(!inq[e.v]) {
cnt[e.v]++;
if(cnt[e.v] >= n) return false;
q.push(e.v);
inq[e.v] = ;
}
}
}
}
return true;
} int main()
{
fast;
int T, kase = ;
cin >> T;
while (T--)
{
cin >> n >> m >> x;
for(int i = ; i <= n; i++)
{
g[i].clear();
dis[i] = 1e18;
cnt[i] = inq[i] = ;
}
for(int i = ; i < n; i++)
{
addedge(i+, i, -);
addedge(i, i+, 2e9);
}
for(int i = ; i < m; i++)
{
int a, b, c, d;
cin >> a >> b >> c >> d;
if(a == b)
{
if(c == d)
{
addedge(a, c, x);
addedge(c, a, -x);
}
else
{
addedge(a, c, x - );
addedge(d, a, -x - );
}
}
else
{
if(c == d)
{
addedge(b, c, x - );
addedge(c, a, -x - );
}
else
{
addedge(b, c, x - );
addedge(d, a, -x - );
}
}
}
cout << "Case #" << (kase++);
if(!spfa()) cout << " IMPOSSIBLE\n";
else {
for(int i = ; i <= n; i++)
cout << ' ' << dis[i] - dis[i - ];
cout << endl;
}
}
return ;
}
Problem G - Alice’s Stamps 02：24：14 （-2） Solved by Dancepted （dp）
上机的时候思路有乱，调出来之后交上去因为N开大了MLE返回CE，贡献了两发罚时。
dfs枚举当前搜索过的最右端点和剩下的k的数量，记忆化一下，状态数最多就$O(n^{2})$。
HDU的T组数据比较玄学，差点没敢交。
代码：$O(n^{2})$
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 2005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x)
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } struct Node{
int l, r;
bool operator < (const Node& x) const {
if (r == x.r) {
return l < x.l;
}
return r < x.r;
}
}; int n, m, k;
vector<Node> ns, ns1;
int ans = ;
int f[N][N];
int dfs(int id, int r, int cnt, int resk) {
if (f[r][resk]) {
return f[r][resk];
}
if (resk == ) {
return ;
}
if (id >= sz(ns)) {
return ;
}
int tmp = dfs(id+, r, cnt, resk);
tmp = max(tmp, ns[id].r - max(ns[id].l-, r) + dfs(id+, ns[id].r, cnt + ns[id].r - max(ns[id].l-, r), resk-));
ans = max(ans, cnt + tmp);
return f[r][resk] = tmp;
}
bool ok[N];
int l[N], r[N];
int main() {
int T, kase = ;
cin >> T;
while (T--) {
ns.clear(), ns1.clear();
read(n, m, k);
for (int i = ; i <= m; i++) {
read(l[i], r[i]);
ok[i] = true;
}
for (int i = ; i <= m; i++) {
for (int j = ; j <= m; j++) if (i != j) {
if (l[i] < l[j] && r[j] <= r[i]) {
ok[j] = false;
}
}
}
for (int i = ; i <= m; i++) if (ok[i]) {
ns.push_back(Node{l[i], r[i]});
}
sort(ns.begin(), ns.end()); // init();
ans = ;
for (int i = ; i <= n; i++) {
for (int j = ; j <= k; j++) {
f[i][j] = ;
}
}
dfs(, , , k);
printf("Case #%d: %d\n", kase++, ans);
}
return ;
}
题解的做法是一个很好写的dp：
$f_{i, j}$表示最右边的覆盖点是i的情况下，用了j个集合的最大覆盖长度。
代码：$O(n^{2})$
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 2005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } int mxr[N];
int f[N][N];
int main() {
int T; cin >> T;
for (int kase = ; kase <= T; kase++) {
int n, m, k; read(n, m, k);
memset(mxr, , (n+) * sizeof(int));
for (int i = ; i <= n; i++) {
for (int j = ; j <= k; j++) {
f[i][j] = ;
}
}
for (int i = ; i <= m; i++) {
int l, r; read(l, r);
mxr[l] = max(mxr[l], r);
}
int ans = , r = ;
for (int i = ; i <= n; i++) {
r = max(mxr[i], r);
for (int j = ; j <= k; j++) {
f[i][j] = max(f[i][j], f[i-][j]);
ans = max(ans, f[i][j]);
if (j+ <= k)
f[r][j+] = max(f[r][j+], f[i-][j] + r - i + ),
ans = max(ans, f[r][j+]);
}
}
printf("Case #%d: %d\n", kase, ans);
}
return ;
}
Problem I - Inkopolis 04：59：59 （-3） Solved by Dancepted & xk
中午还在看的基环树下午就用到了。
先考虑树上的情况。会影响答案的部分只有修改颜色之前的颜色$color_{pre}$，和之后的颜色$color_{now}$两种。看是否会新增、删除color region，或者合并、分割原来的color region。
再考虑环上的情况。同上。特别地：如果整个环都是同一种颜色，修改了颜色不会分割原来的color region；如果整个环除了修改的边都是$color_{now}$，则修改颜色不会合并color region。
计算节日开始前各街道的颜色，可以模仿修改的操作，一条条边加入就行了。
实现的时候先把环扣出来，统计一下环各颜色的边数量。用个map保存每个点相邻的各颜色的数量。再用map保存每条边的颜色。
然后扫一遍所有的边先计算初始情况下的region的数量sum，然后一次次修改边的颜色，更新sum就好了。
代码：O(nlogn)
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 200005
#define M 200005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x) using namespace std;
typedef long long ll;
typedef double db; /** fast read **/
template <typename T>
inline void read(T &x) {
x = ; T fg = ; char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fg = -;
ch = getchar();
}
while (isdigit(ch)) x = x*+ch-'', ch = getchar();
x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
int len = ; char c[]; if (x < ) putchar('-'), x = -x;
do{++len; c[len] = x% + '';} while (x /= );
for (int i = len; i >= ; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); } int tot;
int head[N], nxt[N<<], ver[N<<], col[N<<];
void addEdge(int u, int v, int c) {
nxt[tot] = head[u], ver[tot] = v, col[tot] = c, head[u] = tot++;
} int iscir[N], cirlen;
bool vis[N];
int getCir(int u, int fa, int& rt) {
vis[u] = true;
for (int i = head[u]; i != -; i = nxt[i]) {
int v = ver[i];
if (v == fa) continue;
if (vis[v]) {
rt = v;
iscir[u] = ;
cirlen++;
return ;
}
else {
int tmpcir = getCir(v, u, rt);
if (tmpcir) {
iscir[u] = tmpcir;
cirlen++;
return u != rt;
}
}
}
return ;
} int cnt[N];
int sum;
map<int, int> mp[N];
map<pair<int,int>, int> uv_col; void update(int u, int v, int ccur, bool ifprint) {
if (u > v) swap(u, v);
int cpre = uv_col[mp(u, v)];
if (cpre == ccur) {
if (ifprint)
printf("%d\n", sum);
return;
}
if (!iscir[u] || !iscir[v]) {
//on tree
if (mp[u][cpre] >= && mp[v][cpre] >= ) {
sum++;
}
if (mp[u][cpre] == && mp[v][cpre] == ) {
sum--;
}
if (mp[u][ccur] && mp[v][ccur]) {
sum--;
}
if (!mp[u][ccur] && !mp[v][ccur]) {
sum++;
}
}
else {
//on circle
if (mp[u][cpre] >= && mp[v][cpre] >= ) {
if (cnt[cpre] != cirlen)
sum++;
}
if (mp[u][cpre] == && mp[v][cpre] == ) {
sum--;
}
if (mp[u][ccur] && mp[v][ccur]) {
if (cnt[ccur] != cirlen-)
sum--;
}
if (!mp[u][ccur] && !mp[v][ccur]) {
sum++;
} if (cpre != -)
cnt[cpre]--;
cnt[ccur]++;
}
uv_col[mp(u, v)] = ccur;
if (cpre != -) {
mp[u][cpre]--;
mp[v][cpre]--;
}
mp[u][ccur]++;
mp[v][ccur]++;
if (ifprint)
printf("%d\n", sum);
} int main() {
int T;
cin >> T;
for (int kase = ; kase <= T; kase++) {
int n, m; read(n, m);
// init
uv_col.clear();
for (int i = ; i <= n; i++) {
mp[i].clear();
}
tot = ;
memset(head, -, (n+) * sizeof(int)); // input and count colors on every edge
for (int i = ; i <= n; i++) {
int u, v, c; read(u, v, c);
addEdge(u, v, c);
addEdge(v, u, c);
if (u > v)
swap(u, v);
uv_col[mp(u, v)] = -;
// mp[u][c]++; mp[v][c]++;
}
// get circle
int rt = -;
memset(vis, false, (n+) * sizeof(bool));
memset(iscir, , (n+) * sizeof(int));
cirlen = ;
getCir(, -, rt);
// count colors on circle
memset(cnt, , (n+) * sizeof(int));
sum = ;
for (int i = ; i < n; i++) {
int u = ver[i<<], v = ver[i<<|], ccur = col[i<<];
update(u, v, ccur, false);
} printf("Case #%d:\n", kase);
for (int i = ; i <= m; i++) {
int u, v, ccur; read(u, v, ccur);
update(u, v, ccur, true);
}
}
return ;
}
/*
11111
5 10
1 5 1
2 5 1
3 5 1
4 5 1
1 2 1
1 2 1 2
4 4
1 2 1
2 3 1
3 4 1
4 1 1
1 2 2
3 4 2
2 3 2
4 1 4
4 3
4 2 4
2 3 3
3 4 2
1 4 1
3 4 2
2 3 4
3 4 3
*/
总结：
本场能绝杀还是有点运气成分在的，而且现场赛的话最后几分钟可能就交不上题了也可能。不过绝杀也不影响苟在银牌，问题不大。
然后最近的几场好像都是我在贡献罚时，感觉码力出现了一些小小的问题。这两天要小心一点。
然后还有两周可能就要退役了，模拟赛还没有进过金区感觉很难受啊，打成这个样子还怎么冲金啊qaq。
模拟赛小结：2017 China Collegiate Programming Contest Final (CCPC-Final 2017)的更多相关文章
- The 2017 China Collegiate Programming Contest, Hangzhou Site Solution
A: Super_palindrome 题面:给出一个字符串,求改变最少的字符个数使得这个串所有长度为奇数的子串都是回文串 思路:显然,这个字符串肯定要改成所有奇数位相同并且所有偶数位相同 那统计一下 ...
- 2017 China Collegiate Programming Contest Final (CCPC 2017)
题解右转队伍wiki https://acm.ecnu.edu.cn/wiki/index.php?title=2017_China_Collegiate_Programming_Contest_Fi ...
- The 2015 China Collegiate Programming Contest A. Secrete Master Plan hdu5540
Secrete Master Plan Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Othe ...
- The 2015 China Collegiate Programming Contest Game Rooms
Game Rooms Time Limit: 4000/4000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) Submi ...
- 2016 China Collegiate Programming Contest Final
2016 China Collegiate Programming Contest Final Table of Contents 2016 China Collegiate Programming ...
- 2018 China Collegiate Programming Contest Final (CCPC-Final 2018)-K - Mr. Panda and Kakin-中国剩余定理+同余定理
2018 China Collegiate Programming Contest Final (CCPC-Final 2018)-K - Mr. Panda and Kakin-中国剩余定理+同余定 ...
- 模拟赛小结：The 2019 China Collegiate Programming Contest Harbin Site
比赛链接:传送门 上半场5题,下半场疯狂挂机,然后又是差一题金,万年银首也太难受了. (每次银首都会想起前队友的灵魂拷问:你们队练习的时候进金区的次数多不多啊?) Problem J. Justify ...
- 模拟赛小结：2018 China Collegiate Programming Contest Final (CCPC-Final 2018)
比赛链接:传送门 跌跌撞撞6题摸银. 封榜后两题,把手上的题做完了还算舒服.就是罚时有点高. 开出了一道奇奇怪怪的题(K),然后ccpcf银应该比区域赛银要难吧,反正很开心qwq. Problem A ...
- The 2019 China Collegiate Programming Contest Harbin Site F. Fixing Banners
链接: https://codeforces.com/gym/102394/problem/F 题意: Harbin, whose name was originally a Manchu word ...
随机推荐
- URL encoding(URL编码)
URL encoding(URL编码),也称作百分号编码(Percent-encoding),是指特定上下文的统一资源定位符(URL)编码机制UrlEncode:将字符串以URL编码返回值:字符串函数 ...
- jmeter设置全局变量与正则表达式提取器
接口测试中,很多接口都要带上登录后的token才能正常发送请求,这里记录一下登录获取token设置为全局变量供其他接口使用 登录后返回信息信息中会有一个token值,添加后置处理器中的正则表达式提取t ...
- SparkCore的性能优化
1.广播变量 1.1. Spark提供的Broadcast Variable,是只读的,并且在每个节点上只会有一份副本,而不会为每个task都拷贝一份副本 1.2.它的最大作用,就是减少变量到各个节点 ...
- 《Visual Studio程序员箴言》笔记
还记得刚工作时看到某前辈用快捷键操作Visual Studio,赞叹不已,才发觉原来快捷键熟练了效率可以如此之高.后来,我在为一个经常使用的命令添加快捷键的时候又发现Visual Studio的快捷键 ...
- Pandas使用细则
本文介绍pandas的使用,总结了我在机器学习过程中常使用到的一些方法等. #pandas学习 import pandas as pd import numpy as np import seabor ...
- 【机器学习】深入理解拉格朗日乘子法（Lagrange Multiplier) 和KKT条件
在求取有约束条件的优化问题时,拉格朗日乘子法(Lagrange Multiplier) 和KKT条件是非常重要的两个求取方法,对于等式约束的优化问题,可以应用拉格朗日乘子法去求取最优值:如果含有不等式 ...
- mysql 5.6 后热数据的加载
mysql 5.6 后热数据的加载 转自:http://blog.itpub.net/20892230/viewspace-2127469/ 故障现象:在数据库重启后,碰巧遇到业务高峰期,连接数满,导 ...
- 利用commons-pool2自定义对象池
一.为什么使用对象池 恰当地使用对象池化技术,可以有效地减少对象生成和初始化时的消耗,提高系统的运行效率.commons-pool2是Apache下一个开源的公共资源池.我们可以根据它来快速的建立 ...
- # 模乘（解决乘法取模爆long long）
模乘(解决乘法取模爆long long) 二进制思想,变乘法为多次加法,具体思想跟着代码手算一遍就理解了,挺简单的 ll qmul(ll a,ll b,ll m) { ll ans=0; while( ...
- thinkphp框架部署出现404解决
1:虚拟机配置文件修改: location / { index index.php index.html; if (!-e $request_filename) { rewrite ^/index.p ...