2019 Multi-University Training Contest 9
A.Rikka with Quicksort
前来填坑= =
题解不想再码一遍了，戳这看题解。
思路挺清晰的，推式子需要用到高中数列的技巧，还是有点巧妙。之后分段打表来搞就行。
注意一点就是最终推出来的式子中：
\[f(n)=\frac{n - 1}{(n + 1)(n + 2)}+f(n - 1),n>m
\]
所以我们对于不同的\(m\)，应该从\(m\)开始计算递推，因为\(n-1\geq m\)。(我一开始就是这里被坑了好久)
打表代码如下：
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1000000007;
ll qp(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return ans;
}
int main() {
//ios::sync_with_stdio(false); cin.tie(0);
ll sum = 0;
for(int i = 1; i <= 1000000000; i++) {
sum += 1ll * (i - 1) * qp(1ll * (i + 1) * (i + 2) % MOD, MOD - 2) % MOD;
sum %= MOD;
if(i % 250000 == 0) cout << sum << ',';
}
return 0;
}
AC代码：
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7, base = 250000;
const int biao[] = {0,430276075,367631060,82156064,416495341,103778161,312866495,372614758,864422505,135244213,839083452,75149398,12274495,929472113,296922560,545282131,165736797,207238703,178687671,463030618,45513528,848929864,959617059,911407582,907969740,659095042,209690930,432634376,998842553,549632212,465169763,553024204,265853810,491139524,657766420,388161912,234752773,983353163,689334482,17344532,211535148,748104578,220532609,893786629,218147054,81483556,534793035,959881708,977778800,371118427,59125624,173486302,155307713,743188128,5109642,569236876,504218398,41781867,558032090,466830542,355562371,238124425,78281979,169970521,657514755,310361120,136221681,254472731,734214987,682656054,135048532,20534979,698555650,958079874,874131761,942466030,428951372,301204149,907140402,547633082,249033323,618079471,342474578,819713719,285148375,865954550,953181098,80929740,274605739,895491124,448568388,123805205,338808663,472404161,476020936,196163506,994414699,240239382,853282425,645869452,608885782,498702455,960456812,1423691,56743518,969402887,588202804,789396184,402157044,673238460,662478809,109749470,832425005,992864284,9066205,767346863,829118770,729394799,928753244,867501430,354377916,386809612,318976895,725629204,435655567,131291942,410301260,221244928,159046301,359593982,137135076,523089785,162937071,833886248,928999173,631291292,184431844,842527267,312900794,894396299,756250834,393384865,889234789,635122403,200861698,759211229,996606186,167312697,901660772,865559209,510861323,611120390,972009731,386146829,293148505,38055023,869619483,610471486,424395325,573870351,927422185,273878248,95462189,668729212,135734976,187016670,418685792,505156759,435907069,463951019,124095618,497942800,741119707,828565633,751342140,923109706,489957190,982707589,912092561,316639334,906042258,225957004,12909523,47279004,791698432,415185110,732306601,529986698,306799763,347159022,120648379,776190492,508431604,772672805,494594060,682589755,108542774,557802368,129661807,580786551,240263064,719310714,897535368,372559676,576987022,10234687,317917754,147614861,485649473,461147490,785051077,752522019,680622326,768136403,377595485,460035981,717071900,434044420,937381849,920977703,624151210,596586624,119953887,711446600,983811862,944916225,586679874,188016035,707663439,66584585,436611606,979994948,521406514,913902044,502310459,412389409,975029971,998539050,504025836,278566363,587205928,33556674,410828125,242968511,608133297,573996044,421860420,999429443,296993162,394220938,867622445,96647644,721500554,263031174,655855168,695111461,993238219,431519266,914118475,317251351,51262505,323301407,589135201,112395006,71578037,429705756,247240521,106637126,360906467,491004779,594945904,33463242,184731867,80889757,805760293,831050394,833028231,686520512,910357398,693116921,82224834,756552235,975660951,594266001,268408365,720155309,847449717,650583739,146356515,987759346,562620479,77703814,929728098,537673876,852004855,758918848,206473409,432003382,946394676,435562245,863220082,97939517,649322791,461140689,280428957,161242774,357678382,85908285,226748229,664229757,807653212,450115089,877487170,918238788,778693149,72817560,994174733,880374288,945894295,37595677,27648222,534300197,931682819,911395094,808783571,429882934,635791874,190617949,74852114,825127333,550341803,343365291,775821876,4236874,921901579,864694234,135151256,386713078,377664017,93035598,927497455,105307879,398306653,740225505,765008343,257047250,959715351,448603188,887165205,867427501,108229174,433415600,519437663,265577808,885845356,825251947,957846611,231305947,630517004,118126575,528227783,740683510,231078471,68992178,414230982,344400899,53219084,724736420,666735328,12104846,768839578,990121728,597180915,250870380,609184788,517652538,42744559,126393448,914722723,647067478,380932379,337266904,271768052,444799367,348308161,279469291,320228531,96507612,492753477,586197744,393150554,402687584,583552826,379190699,684709488,418863144,53005468,172741582,972949808,296586001,846505392,929476843,286515437,373984842,685010707,444942883,276211775,904128865,383307622,325276416,177895317,293546816,56790451,257386998,743152811,883231752,684014983,173870794,435380032,968483153,340540340,329230477,129698818,821278359,621001948,971036594,170167081,118656519,816303098,658091678,871518425,43270904,29935919,584192702,88492245,815508333,73162324,213380044,438852764,260979292,967960859,70989479,855761802,945337755,46102073,707240065,42991241,838021956,899095549,628310720,839358117,197102910,991028554,58802467,648713407,622315385,346421501,487692702,199916211,814337702,502873189,235396696,37986776,941524078,478280026,294678109,924420678,213394313,600984357,24389351,973132888,981199442,49449177,220502353,287659969,685522669,659480720,954473502,260894110,821294025,313954869,256989151,419149432,902768142,302216867,916023211,66645682,741140752,651621754,424645267,232509196,337989297,545261467,720537822,81151904,451310447,451666784,517572146,782505112,868327896,229887728,208315451,214742552,302477527,245148458,82971608,18460011,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int n, m, T;
ll qp(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return ans;
}
int calc(int k) {
int tmp = k / base;
int ans = biao[tmp];
for(int i = base * tmp + 1; i <= k; i++) {
ans = (ans + 1ll * (i - 1) * qp(1ll * (i + 1) * (i + 2) % MOD, MOD - 2) % MOD) % MOD;
}
return ans;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> T;
while(T--) {
cin >> n >> m;
if(n <= m) {
cout << 0 << '\n';
continue;
}
int ans = ((calc(n - 1) - calc(m)) % MOD + MOD ) % MOD;
ans = 1ll * ans * (n + 1) % MOD * n % MOD;
ans = (n - 1 + 2ll * ans * qp(n, MOD - 2) % MOD) % MOD;
cout << ans << '\n';
}
return 0;
}
B.Rikka with Cake
题意：
现在有一块大小为\(n*m\)的蛋糕，并且会切\(k\)刀，对于每一刀，都会有\((x,y,d)\)来表示起点位于\((x,y)\)，方向为\(d\)一直到边界的一刀。
现在问最后蛋糕会被分为多少块。
思路：
手玩一下样例即可发现答案为交点数\(+1\)。
扫描线+树状数组统计即可。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int T;
int n, m, k;
int b[N], c[N];
struct point{
int x, y;
char d;
}a[N];
void Hash() {
sort(b + 1, b + b[0] + 1);
sort(c + 1, c + c[0] + 1);
b[0] = unique(b + 1, b + b[0] + 1) - b - 1;
c[0] = unique(c + 1, c + c[0] + 1) - c - 1;
for(int i = 1; i <= k; i++) {
a[i].y = lower_bound(b + 1, b + b[0] + 1, a[i].y) - b;
a[i].x = lower_bound(c + 1, c + c[0] + 1, a[i].x) - c;
}
}
int p[N];
int lowbit(int x) {
return x & -x;
}
void upd(int x, int v) {
for(; x < N; x += lowbit(x)) p[x] += v;
}
int query(int l, int r) {
if(l > r) return 0;
int ans = 0;
for(int i = r; i; i -= lowbit(i)) ans += p[i];
for(int i = l; i; i -= lowbit(i)) ans -= p[i];
return ans;
}
vector <int> v[N], add[N], del[N];
int L[N], R[N], U[N], D[N];
int main() {
cin >> T;
while(T--) {
scanf("%d%d%d", &n, &m, &k);
b[0] = c[0] = 0;
for(int i = 1; i <= k; i++) {
int x, y; char ch;
scanf("%d %d %c", &a[i].x, &a[i].y, &a[i].d);
b[++b[0]] = a[i].y; c[++c[0]] = a[i].x;
}
Hash();
for(int i = 1; i <= c[0]; i++) add[i].clear(), del[i].clear(), v[i].clear(), U[i] = N, D[i] = 0;
for(int i = 1; i <= b[0]; i++) L[i] = 0, R[i] = N;
ll ans = 1;
for(int i = 1; i <= k; i++) {
if(a[i].d == 'L') {
L[a[i].y] = max(L[a[i].y], a[i].x);
}
if(a[i].d == 'R') {
R[a[i].y] = min(R[a[i].y], a[i].x);
}
if(a[i].d == 'U') {
U[a[i].x] = min(U[a[i].x], a[i].y);
}
if(a[i].d == 'D') {
D[a[i].x] = max(D[a[i].x], a[i].y);
}
}
for(int i = 1; i <= k; i++) {
if(a[i].d == 'U' || a[i].d == 'D') v[a[i].x].push_back(i);
else {
if(a[i].d == 'L') {
if(a[i].x == L[a[i].y]) {
add[1].push_back(i);
del[a[i].x].push_back(i);
}
} else {
if(a[i].x == R[a[i].y]) {
add[a[i].x].push_back(i);
del[c[0]].push_back(i);
}
}
}
}
for(int i = 1; i <= c[0]; i++) {
for(auto it : add[i]) {
upd(a[it].y, 1);
}
for(auto it : v[i]) {
char d = a[it].d;
if(d == 'U' && a[it].y == U[i]) {
ans += query(a[it].y - 1, N - 1);//[y,N-1]
} else if(d == 'D' && a[it].y == D[i]){
ans += query(0, a[it].y);//[1,y]
}
}
for(auto it : del[i]) {
upd(a[it].y, -1);
}
}
printf("%lld\n", ans);
}
return 0;
}
C.Rikka with Mista
题意：
给出\(n\)个数\((n\leq 40)\)，每个数都有取或者不取两种情况，之后将取了的数加起来。
现在问所有的\(2^n\)种情况中，\(4\)出现的次数为多少。
思路：
- 看到数据范围，考虑meet in the middle，折半枚举情况。
- 直接算显然是不可能的，考虑每一位的贡献。
- 怎么考虑？
- 取模运算即可。对于\(a_i,b_j\)，如果\((a_i+b_j)\% 10^k=4\)，那么它们在\(k\)位有贡献。
- 容易想到要对后缀排序，之后就能二分加速，或者直接two pointer。
- 每次排序显然复杂度过高，因为到我们是按位考虑贡献的，所以...
- 基数排序！每次可以优化一个\(log\)。
总时间复杂度为\(O(100n)\)。
Code
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll, int> pli;
const int N = 45, MAX = (1 << 20) + 5;
int T, n;
int a[N];
vector <pli> x, y, A[10], B[10];
void solve(int pos, int r, ll sum, int op) {
if(pos == r) {
if(op) x.push_back(MP(sum, 0));
else y.push_back(MP(sum, 0));
return;
}
solve(pos + 1, r, sum, op);
solve(pos + 1, r, sum + a[pos + 1], op);
}
ll work1(int k1, int k2, ll lim) {
ll res = 0;
int now = 0, sz = B[k2].size();
for(int i = A[k1].size() - 1; i >= 0; i--) {
while(now < sz && A[k1][i].se + B[k2][now].se < lim) now++;
res += now;
}
return res;
}
ll work2(int k1, int k2, ll lim) {
ll res = 0;
int sz = B[k2].size(), now = sz - 1;
for(int i = 0; i < A[k1].size(); i++) {
while(now >= 0 && B[k2][now].se + A[k1][i].se >= lim) now--;
res += sz - now - 1;
}
return res;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> T;
while(T--) {
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
int mid = n / 2;
x.clear(), y.clear();
solve(0, mid, 0, 0); solve(mid, n, 0, 1);
ll base = 1, ans = 0;
for(int t = 1; t <= 11; t++) {
for(int i = 0; i < 10; i++) A[i].clear(), B[i].clear();
for(auto it : x) A[it.fi % 10].push_back(MP(it.fi / 10, it.se));
for(auto it : y) B[it.fi % 10].push_back(MP(it.fi / 10, it.se));
for(int i = 0; i < 10; i++) {
int k1 = (14 - i) % 10, k2 = (13 - i) % 10;
ans += work1(i, k1, base);
ans += work2(i, k2, base);
}
int now = 0;
for(int i = 0; i < 10; i++) {
ll tmp = i * base;
for(auto it : A[i]) {
x[now++] = MP(it.fi, tmp + it.se);
}
}
now = 0;
for(int i = 0; i < 10; i++) {
ll tmp = i * base;
for(auto it : B[i]) {
y[now++] = MP(it.fi, tmp + it.se);
}
}
base *= 10;
}
cout << ans << '\n';
}
return 0;
}
E.Rikka with Game
题意：
给出一个字符串，两个人在玩博弈游戏。两个人依次执行操作，要么中止游戏，要么将某个数往后移动一位：\(a->b,\cdots,z->a\)。
现在第一个人想要字符串最小，而第二个人想要字符串最大。
问最终局面是什么。
思路：
贪心考虑：
- 对于第一个人来说，目前局面不存在\(z\)或者在第一个\(z\)之前有某个字符小于\(y\)，那么他就应该立即中止游戏；
- 同理，对于第二个人来说，如果目前字符全不小于\(y\)，或者在一个较小字符前面存在\(z\)，那么他就应该立即终止游戏；
- 其余情况，贪心考虑即可。
代码如下：
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int T;
const int N = 105;
char s[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> T;
while(T--) {
cin >> s + 1;
int f = 0;
int n = strlen(s + 1);
while(1) {
f = 0;
for(int i = 1; i <= n; i++) {
if(s[i] == 'z') {
s[i] = 'a';
f = 1; break;
} else if(s[i] < 'y') break;
}
if(!f) {
cout << s + 1 << '\n';
break;
}
for(int i = 1; i <= n; i++) {
if(s[i] < 'y') {
s[i]++;
f = 2;
break;
} else if(s[i] == 'z') {
break;
}
}
if(f < 2) {
cout << s + 1 << '\n';
break;
}
}
}
return 0;
}
F.Rikka with Coins
题意：
现在有面值为\(10,20,50,100\)的纸币，每种都有无限多张。
现在有\(n\)个物品，每种物品都有价值\(a_i\)，问最少带多少张纸币，对于每个物品都能刚好给够。
思路：
简单分析就会发现：
- \(10\)元纸币最多一张；
- \(20\)元纸币最多四张；
- \(50\)元纸币最多一张。
那么就直接暴力枚举前三种面额多少张即可，\(100\)元纸币的情况可以直接计算出来。
Code
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 105;
int T, n;
int w[N];
int main() {
cin >> T;
while(T--) {
cin >> n;
for(int i = 1; i <= n; i++) cin >> w[i];
int ans = INF;
for(int i = 1; i <= n; i++) {
if(w[i] % 10 != 0) ans = -1;
}
if(ans == -1) {
cout << ans << '\n';
continue;
}
int f;
for(int a = 0; a <= 1; a++) {
for(int b = 0; b <= 4; b++) {
for(int c = 0; c <= 1; c++) {
int sum, mx, tmp = 0;
for(int i = 1; i <= n; i++) {
f = 1; mx = INF;
for(int aa = 0; aa <= a; aa++) {
for(int bb = 0; bb <= b; bb++) {
for(int cc = 0; cc <= c; cc++) {
sum = aa * 10 + bb * 20 + cc * 50;
if(w[i] >= sum && (w[i] - sum) % 100 == 0) {
mx = min(mx, (w[i] - sum) / 100);
}
}
}
}
if(mx == INF) {
f = 0; break;
} else tmp = max(tmp, mx);
}
if(f) ans = min(ans, a + b + c + tmp);
}
}
}
if(ans == INF) ans = -1;
cout << ans << '\n';
}
return 0;
}
G.Rikka with Travels
题意：
定义\(L(a,b)\)为一棵树上从\(a\)到\(b\)经过的点的数量。
求出所有的点对\((L(a,b),L(c,d))\)并且满足\(a->b\)不与\(c->d\)相交。
思路：
比较直接的想法就是换根树\(dp\)。
另外还有比较巧妙的想法：树的直径。
因为树的直径具有最长性，那么：
- 对于点对\((a,b)\)，如果\(a\)没在直径上，那么\(b\)一定是整个直径；
- 如果路径\(a,b\)都在直径上面，最优情况肯定是它们的一端都在直径的某一端（最长性）；
- 如果都没在直径上面，这种情况一定可以被前两种情况覆盖。
形象来说，就是考虑如下两种情况：
有一点需要注意的是，最后得到答案的时候要倒着扫一遍才能得到所有的\(ans\)。
详见代码：
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int T;
int n;
struct Edge{
int v, next;
}e[N << 1];
int head[N], tot;
void adde(int u, int v) {
e[tot].v = v; e[tot].next = head[u]; head[u] = tot++;
}
int d[N], fa[N];
void dfs(int u, int pa) {
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].v;
if(v == pa) continue;
d[v] = d[u] + 1; fa[v] = u;
dfs(v, u);
}
}
bool chk[N];
int f[N], g[N];
void dfs2(int u, int pa) {
f[u] = g[u] = 1;
for(int i = head[u]; i != -1; i = e[i].next) {
int v = e[i].v;
if(v == pa || chk[v]) continue;
dfs2(v, u);
g[u] = max(g[u], f[u] + f[v]);
f[u] = max(f[u], f[v] + 1);
}
}
int L[N], R[N], ans[N];
vector <int> l;
void upd(int k1, int k2) {
ans[k1] = max(ans[k1], k2);
ans[k2] = max(ans[k2], k1);
}
void work1() {
int sz = l.size();
for(int i = 0; i < sz; i++) {
L[i] = f[l[i]] + i;
R[i] = sz - i + f[l[i]] - 1;
}
for(int i = 1; i < sz; i++) L[i] = max(L[i], L[i - 1]);
for(int i = 1; i < sz; i++) {
upd(L[i - 1], R[i]);
}
}
void work2() {
for(int i = 1; i <= n; i++) {
if(!chk[i]) upd(g[i], (int)l.size());
}
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> T;
while(T--) {
memset(head, -1, sizeof(head)); tot = 0;
cin >> n;
for(int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
adde(u, v); adde(v, u);
}
d[1] = 0; dfs(1, 0);
int u = 1;
for(int i = 1; i <= n; i++) if(d[i] > d[u]) u = i;
d[u] = 0; dfs(u, 0);
int v = u;
for(int i = 1; i <= n; i++) if(d[i] > d[v]) v = i;
memset(chk, 0, sizeof(chk));
chk[u] = 1; l.clear();
for(int i = v; i != u; i = fa[i]) chk[i] = 1, l.push_back(i);
l.push_back(u);
for(int i = 1; i <= n; i++) if(chk[i]) dfs2(i, 0);
work1(); work2();
for(int i = n - 1; i >= 1; i--) ans[i] = max(ans[i], ans[i + 1]);
ll res = 0;
for(int i = 1; i <= n; i++) res += ans[i], ans[i] = 0;
cout << res << '\n';
}
return 0;
}
H.Rikka with Stable Marriage
题意：
给出\(a_i\)和\(b_i\),求其两两匹配异或和的最大值。
思路：
其实原题求的是稳定婚姻匹配问题，但可以将问题转化为这个。
因为贪心匹配过后，就不能再交换伴侣了，剩下的在一起也只能这样了。想找更优秀的也没法= =优秀的可能早就和别人在一起了。就算优秀的跑来找你，那另一个人也会剩下来，所以何必呢。（口胡）
解法就类似于传送门中的\(B\)题。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 2;
int T, n;
int a[N], b[N];
inline char gc() {
static char buf[131072], *p1, *p2;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 131072, stdin), p1 == p2) ? EOF : *p1++;
}
int read() {
int x = 0;
char c = gc();
for (; c < 48 || c > 57; c = gc())
;
for (; c > 47 && c < 58; x = x * 10 + c - 48, c = gc())
;
return x;
}
struct Trie{
int ch[N * 30][2], cnt[N * 30];
int tot;
void init() {
tot = 0;
ch[tot][0] = ch[tot][1] = 0;
}
int New_node() {
ch[++tot][0] = ch[tot][1] = 0;
return tot;
}
void Insert(int x) {
int u = 0;
for(register int i = 29; ~i; --i) {
int p = ((x >> i & 1) ? 1 : 0);
if(!ch[u][p]) ch[u][p] = New_node();
u = ch[u][p];
++cnt[u];
}
}
}t1, t2;
ll work() {
int u1, u2;
ll ans = 0;
while(n--) {
u1 = u2 = 0;
int x = 0;
for(register int p = 29; ~p; --p) {
if(t1.cnt[t1.ch[u1][0]] && t2.cnt[t2.ch[u2][1]]) {
--t1.cnt[t1.ch[u1][0]];
--t2.cnt[t2.ch[u2][1]];
u1 = t1.ch[u1][0]; u2 = t2.ch[u2][1];
x ^= (1 << p);
}else if(t1.cnt[t1.ch[u1][1]] && t2.cnt[t2.ch[u2][0]]){
--t1.cnt[t1.ch[u1][1]];
--t2.cnt[t2.ch[u2][0]];
u1 = t1.ch[u1][1]; u2 = t2.ch[u2][0];
x ^= (1 << p);
}else if(t1.cnt[t1.ch[u1][0]] && t2.cnt[t2.ch[u2][0]]) {
--t1.cnt[t1.ch[u1][0]];
--t2.cnt[t2.ch[u2][0]];
u1 = t1.ch[u1][0]; u2 = t2.ch[u2][0];
} else {
--t1.cnt[t1.ch[u1][1]];
--t2.cnt[t2.ch[u2][1]];
u1 = t1.ch[u1][1]; u2 = t2.ch[u2][1];
}
}
ans += x;
}
return ans;
}
int main() {
//ios::sync_with_stdio(false); cin.tie(0);
T = read();int i;
while(T--) {
n = read();
for(i = 1; i <= n; ++i) a[i] = read();
for(i = 1; i <= n; ++i) b[i] = read();
t1.init(); t2.init();
for(i = 1; i <= n; ++i) t1.Insert(a[i]);
for(i = 1; i <= n; ++i) t2.Insert(b[i]);
printf("%lld\n", work());
}
return 0;
}
P.S:dfs版本感觉写起来挺容易的，就是时间稍微有点慢，也放出来吧：
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int T, n;
int a[N], b[N];
int tot;
int rt[2], ls[N * 70], rs[N * 70], cnt[N * 70];
int newnode() {
ls[++tot] = rs[tot] = cnt[tot] = 0;
return tot;
}
void insert(int &o, int v, int d) {
if(!o) o = newnode();
cnt[o]++;
if(d < 0) return;
if(v >> d & 1) {
insert(rs[o], v, d - 1);
} else insert(ls[o], v, d - 1);
}
ll query(int o1, int o2, int d, int w) {
if(!cnt[o1] || !cnt[o2]) return 0;
ll res = 0;
if(d == -1) {
int num = min(cnt[o1], cnt[o2]);
cnt[o1] -= num; cnt[o2] -= num;
res += 1ll * w * num;
return res;
}
res = query(ls[o1], rs[o2], d - 1, w + (1 << d)) + query(rs[o1], ls[o2], d - 1, w + (1 << d));
res += query(ls[o1], ls[o2], d - 1, w) + query(rs[o1], rs[o2], d - 1, w);
cnt[o1] = cnt[rs[o1]] + cnt[ls[o1]];
cnt[o2] = cnt[rs[o2]] + cnt[ls[o2]];
return res;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> T;
while(T--) {
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
for(int i = 1; i <= n; i++) cin >> b[i];
rt[0] = rt[1] = tot = 0;
for(int i = 1; i <= n; i++) insert(rt[0], a[i], 31);
for(int i = 1; i <= n; i++) insert(rt[1], b[i], 31);
cout << query(rt[0], rt[1], 31, 0) << '\n';
}
return 0;
}
重要的是自信，一旦有了自信，人就会赢得一切。
