Problem A Mex Query
簽到題
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[maxn];
int main(){
int T;
scanf("%d", &T);
while(T--){
int n;
clr(a);
set<int> se;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d", &a[i]);
se.insert(a[i]);
}
set<int>:: iterator it;
int i = 0;
bool flag = false;
for(it = se.begin(); it != se.end(); it++){
if(i != *it){
printf("%d\n", i);
flag = true;
break;
}
i++;
}
if(!flag) printf("%d\n", i);
}
return 0;
}
問題 B: icebound的商店
DP一下
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 9;
ll a[MAXN], b[MAXN];
ll ans = 0, n;
ll dp[20][3006];
int main(){
a[1] = 1, a[2] = 2, a[3] = 3;
for(int i = 4; i <= 15; i++)
a[i] = a[i-1] + a[i-2];
int t;
scanf("%d", &t);
while(t--) {
clr(dp);
for(int i = 1; i <= 15; i++)
dp[i][0] = 1;
scanf("%lld", &n);
for(int i = 1; i <= 15; i++){
for(int j = 1; j <= n; j++){
dp[i][j] = dp[i-1][j];
if(j >=a [i])
dp[i][j] = (dp[i-1][j] + dp[i][j-a[i]]) % MOD;
}
}
printf("%lld\n", dp[15][n]);
}
return 0;
}
問題 C: Nim Game
區間的nim博弈， 清空數組用多少清多少就沒超時。我看也有人把2^n預處理出來也沒超時，這種處理emmmm能過就行吧。
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ll n, m;
ll tag[MAXN], a[MAXN];
ll qkm(ll a, ll b, ll mode){
ll sum = 1;
while (b) {
if (b & 1) {
sum = (sum * a) % mode;
b--;
}
b /= 2;
a = a * a % mode;
}
return sum;
}
int main(){
int _;
scanf("%d", &_);
while(_--){
//clr(tag); clr(a);
scanf("%lld%lld", &n, &m);
for(int i = 1; i <= n; i++){
scanf("%lld", &a[i]);
tag[i] = tag[i-1] ^ a[i];
}
ll l, r, ans = 0;
for (int i = 1; i <= m; ++i){
scanf("%lld%lld", &l, &r);
ll num = tag[r] ^ tag[l-1];
if(num != 0){
ll temp = qkm(2, m-i, MOD);
ans = (ans + temp) % MOD;
}
}
printf("%lld\n", ans);
for(int i = 0; i <= n; i++){
a[i] = tag[i] = 0;
}
}
return 0;
}
問題 D: Defending Plan Support
問題 E: Bitmap
問題 F: 神殿
水題，從l的最低位向高位開始變化，遇到0變成1 判斷是否小於r
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll z[100];
int main(){
ll l, r;
scanf("%lld%lld", &l, &r);
ll k = l;
ll cnt = 0, num = 0;
while(k){
z[cnt] = k % 2;
k /= 2;
cnt++;
}
ll ans = l,w=1;
for(ll i = 0; i < 65; i++){
if(z[i] == 0){
ll sum = ans + ((ll)1<<i);
if(sum <= r){
z[i] = 1;
num++;
ans = sum;
}
else break;
}
}
printf("%lld\n",ans);
return 0;
}
問題 G: K Multiple Longest Commom Subsequence
問題 H: 跑圖
bfs一遍標記
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 505;
int mp[N][N];
bool vis[N][N];
int n, m, x;
int dir[2][4] = {{0, 0, 1, -1}, {1, -1, 0, 0}};
struct node{
int x, y, t;
};
void bfs(int x, int y){
clr(vis);
queue<node> Q;
node u;
u.x = x; u.y = y; u.t = 0;
Q.push(u);
vis[x][y] = true;
while(!Q.empty()){
node top = Q.front();
Q.pop();
for(int i = 0; i < 4; i++){
int dx = top.x + dir[0][i];
int dy = top.y + dir[1][i];
if(dx>=1 && dx<=n && dy>=1 && dy<=m && top.t+1<=mp[dx][dy] && vis[dx][dy]==0){
node v;
v.x = dx; v.y = dy; v.t = top.t + 1;
mp[dx][dy] = v.t;
vis[dx][dy] = true;
Q.push(v);
}
}
}
}
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
mp[i][j] = INF;
scanf("%d", &x);
if(x == 1) mp[i][j] = 0;
}
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
if(mp[i][j] == 0){
bfs(i, j);
}
}
}
for(int i = 1; i <= n; i++){
for(int j = 1; j < m; j++){
printf("%d ", mp[i][j]);
}
printf("%d\n", mp[i][m]);
}
return 0;
}
/*
2 3
0 0 0
1 0 1
*/
問題 I: Power Seq
問題 J: Beautiful Array
問題 K: 520
簽到題
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 20180520;
ll qkm(ll a, ll b, ll mode)
{
ll sum = 1;
while (b) {
if (b & 1) {
sum = (sum * a) % mode;
b--;
}
b /= 2;
a = a * a % mode;
}
return sum;
}
int main(){
ll n;
scanf("%lld", &n);
ll ans = qkm(2, n, MOD);
printf("%lld\n", ans);
return 0;
}
問題 L: icebound的賬單
簽到題
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a, 0, sizeof(a))
#define line cout<<"------------"<<endl
typedef long long ll;
const int maxn = 2e5 + 10;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int main(){
int n;
cin >> n;
ll sum = 0, x;
for(int i = 0; i < n; i++){
scanf("%lld", &x);
sum += x;
}
if(sum > 0) printf("icebound is happy.\n" );
else if(sum == 0) printf("icebound is ok.\n");
else printf("icebound is sad.\n");
return 0;
}