哈尔滨理工大学软件与微电子学院程序设计竞赛（同步赛）(AK题解)
哈尔滨理工大学软件与微电子学院程序设计竞赛（同步赛）(AK题解)
A-Race
思路：
按照题意模拟一下即可。
代码：
int v1, v2, t, s, l;
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
cin >> v1 >> v2 >> t >> s >> l;
int now1 = 0;
int now2 = 0;
int ghs = 0;
int id;
repd(i, 1, 10000)
{
if (ghs > 0)
{
ghs--;
} else
{
now1 += v1;
}
now2 += v2;
if (now1 >= l || now2 >= l) {
id = i;
break;
}
if (ghs == 0 && now1 - now2 >= t)
{
ghs = s;
}
}
if (now1 == now2 && now2 == l)
{
printf("Tie %d\n", id );
} else if (now1 > now2)
{
printf("Ming %d\n", id );
} else
{
printf("Hong %d\n", id );
}
return 0;
}
B-Min Value
思路：
很老的一种题目了，将数组排序一下，然后对于每一个数\(a_i\)，二分找下数组中\(-1*a_i\)左右两边的下标，然后在这个长度不大于3的小区间里找到\(a_j\)使$ a_i + a_j$的绝对值最小。
代码：
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
pii a[maxn];
int b[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readint();
repd(i, 1, n)
{
a[i].fi = readint();
a[i].se = i;
}
sort(a + 1, a + 1 + n);
repd(i, 1, n)
{
b[i] = a[i].fi;
}
int ans1 = inf;
int ans2 = inf;
repd(i, 1, n)
{
int pos = lower_bound(b + 1, b + 1 + n, -b[i]) - b - 1;
repd(j, max(1, pos), min(n, pos + 3))
{
if (a[i].se == a[j].se)
continue;
int num = abs(b[i] + b[j]);
if (num < ans1)
{
ans1 = num;
ans2 = a[i].se + a[j].se;
} else if (num == ans1)
{
ans2 = min(ans2, a[i].se + a[j].se);
}
}
}
printf("%d %d\n", ans1, ans2 );
return 0;
}
C-Coronavirus
思路：
非常基础的BFS题目，先将地图中不能走的位置处理一下，然后bfs即可。
代码：
const int maxn = 110;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
const int M = maxn;
int n;
int m;
char s[maxn][maxn];
struct BFSer
{
const int dx[4] = { -1, 0, 0, 1}, dy[4] = {0, -1, 1, 0};
int Dis[M][M]; bool vis[M][M]; int Frx[M][M], Fry[M][M];
struct node {int x, y;} Q[M * M];
int l, r;
void bfs(int sx, int sy)
{
for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)Dis[i][j] = 5e8;
Dis[sx][sy] = 0; vis[sx][sy] = true; Q[r++] = (node) {sx, sy};
while (l < r)
{
node t = Q[l++];
int x = t.x, y = t.y;
for (int d = 0; d < 4; d++)
{
int xx = x + dx[d], yy = y + dy[d];
if (xx < 1 || xx > n || yy < 1 || yy > m || vis[xx][yy] || s[xx][yy] == '*')continue;
Dis[xx][yy] = Dis[x][y] + 1; vis[xx][yy] = true;
Frx[xx][yy] = x; Fry[xx][yy] = y;
Q[r++] = (node) {xx, yy};
}
}
}
} bfser;
int dx8[8] = {1, 0, -1, 0, 1, -1, 1, -1};
int dy8[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readint();
m = readint();
repd(i, 1, n)
{
scanf("%s", s[i] + 1);
}
int sx, sy, ex, ey;
repd(i, 1, n)
repd(j, 1, m)
{
if (s[i][j] == 'S')
{
sx = i;
sy = j;
} else if (s[i][j] == 'E')
{
ex = i;
ey = j;
}
}
std::vector<pii> v;
repd(i, 1, n)
{
repd(j, 1, m)
{
if (s[i][j] == '*')
{
v.pb(mp(i, j));
}
}
}
for (auto &X : v)
{
int i = X.fi;
int j = X.se;
repd(k, 0, 7)
{
s[i + dx8[k]][j + dy8[k]] = '*';
}
}
if (s[sx][sy] == 'S')
{
bfser.bfs(sx, sy);
if (bfser.Dis[ex][ey] > 1e8)
{
printf("Impossible\n");
} else
{
printf("%d\n", bfser.Dis[ex][ey]);
}
} else
{
printf("Impossible\n");
}
return 0;
}
D-Array
思路：
这题是Codeforces Round #628 (Div. 2)- D 题的弱化版，
题解可以看：
https://www.cnblogs.com/qieqiemin/p/13163717.html
我交了2种写法，都过了，分别是直接分类讨论构造，还有二进制拆分，都贴上吧。
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
ll x, y;
while (~scanf("%lld %lld", &x, &y))
{
if (x > y)
{
printf("-1\n");
} else
{
if (x == y) {
if (x == 0)
printf("0\n");
else
{
printf("1\n");
// printf("%lld\n", x);
}
} else {
ll c = y - x;
if (c & 1)
{
printf("-1\n");
} else
{
ll z = c / 2;
// chu(z);
// chu(z & x);
if ((z & x) == 0)
{
printf("2\n");
//printf("%lld %lld\n", x + c / 2, c / 2 );
} else
{
printf("3\n");
//printf("%lld %lld %lld\n", x, c / 2, c / 2 );
}
}
}
}
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int a[30];
int b[30];
int c[30];
void PRINF2(ll num, int k)
{
for (int i = k; i >= 0; i--)
{
cout << (bool)(num & (1ll << i));
}
cout << endl;
}
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
int x, y;
while (~scanf("%d %d", &x, &y))
{
// PRINF2(x, 20);
// PRINF2(y, 20);
if (x % 2 != y % 2)
{
printf("-1\n");
} else if (x > y)
{
printf("-1\n");
} else
{
int cnt1 = 0;
int cnt2 = 0;
while (x > 0)
{
a[cnt1++] = x % 2;
x /= 2;
}
while (y > 0)
{
b[cnt2++] = y % 2;
y /= 2;
}
int ans = 0;
for (int i = cnt2; i >= 0; --i)
{
c[i] += b[i];
int flag = 0;
if (b[i] == 1 && a[i] == 0)
{
if (c[i] % 2 == 1)
{
flag = 1;
}
} else if (b[i] == 0 && a[i] == 1)
{
if (c[i] % 2 == 0)
{
flag = 1;
}
} else if (b[i] == 1 && a[i] == 1)
{
if (c[i] % 2 == 0)
{
flag = 1;
}
} else
{
if (c[i] % 2 == 1)
{
flag = 1;
}
}
if (flag)
{
if (i == 0) {
ans = -1;
break;
}
int add = 0;
if (c[i] == 0)
{
for (int j = i + 1; j <= cnt2; ++j)
{
if (c[j] >= 2)
{
c[j] -= 2;
add = 1 << (j - i + 1);
break;
}
}
}
c[i] += add;
c[i]--;
c[i - 1] += 2;
}
}
if (ans == -1)
{
printf("%d\n", -1 );
} else
{
repd(i, 0, 29) {ans = max(ans, c[i]);}
ans %= 4;
printf("%d\n", ans );
}
}
// for (int i = 20; i >= 0; --i) {cout << c[i];}; cout << endl;
repd(i, 0, 29) {a[i] = b[i] = c[i] = 0;}
}
return 0;
}
E-Prize
思路：
对于数字0~9，用bitset<805> 表示数字可以放在中奖号码的哪些位。
然后1~n遍历彩民的号码，
用bitset<805> pre代表上一个位置结束时能匹配到中奖号码的哪些位，
用bitset<805> b代表遍历到当前数字时能匹配到中奖号码的哪些位置。
如果\(b[m]=1\),说明当前位置可以作为一个中奖号码的结束，
同时用数组\(dp_i\)代表以第\(\mathit i\)个数字为结束时，最多中奖多少次。
答案即为\(dp_n\)。
代码:
const int maxn = 3000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n, m;
char s[maxn];
bitset<805> a[11];
bitset<805> b, pre;
int dp[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readint();
m = readint();
scanf("%s", s + 1);
repd(i, 1, m)
{
int num = readint();
while (num--)
{
int x = readint();
a[x][i] = 1;
}
}
pre[0] = 1;
repd(i, 1, n)
{
b = a[s[i] - '0'];
b &= pre << 1;
if (b[m] == 1)
{
dp[i] = dp[i - m] + 1;
} else
{
dp[i] = dp[i - 1];
}
pre = b;
pre[0]=1;
}
if (dp[n] == 0)
{
printf("Failed to win the prize\n");
} else
{
printf("%d\n", dp[n] );
}
return 0;
}
F-Animal Protection
思路：
是单调栈的经典题目，也是一个原题的更改版本，详见：
https://www.cnblogs.com/qieqiemin/p/13164440.html
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 1
bool mpa[3005][3005];
ll height[3005];
ll STK[3005], ANS[3005], head;
char s[1005][1005];
const ll mod = 1000000007;
int main()
{
int n, m;
n = readint();
m = readint();
repd(i, 1, n) {
scanf("%s", s[i] + 1);
}
repd(i, 1, n)
{
repd(j, 1, m)
{
if (s[i][j] == 'X')
{
mpa[i][j] = 0;
} else
{
mpa[i][j] = 1;
}
}
}
long long ans = 0;
for (int i = 1; i <= n; i ++)
{
for (int j = 1; j <= m; j ++)
{
if ( !mpa[i][j] )
height[j] = i;
while ( head and height[ STK[head] ] < height[j] )
{
head --;
}
STK[ ++ head ] = j;
ANS[ head ] = ANS[ head - 1 ] + (i - height[STK[head]] ) * ( STK[head] - STK[head - 1] );
ans += ANS[ head ];
ans %= mod;
}
head = 0;
}
cout << ans;
return 0;
}
G-XOR
思路：
对于这个数列一定可以找到这个数列的最大值\(\mathit n\)在二进制表示法中最高位之后的所有0位置都为1的数字\(num\)
答案即为\(num\oplus n\)，即最大值\(\mathit n\)在二进制表示法中最高位以及最高位之后的所有位都为1。
代码：
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
ll n;
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readll();
if (n == 1)
{
printf("%lld\n", 0ll );
return 0;
}
ll x = 0ll;
while (n > 0)
{
x++;
n /= 2;
}
chu(x);
printf("%lld\n", (1ll << (x )) - 1 );
return 0;
}
H-Maze
思路：
对于整个迷宫看成图，每个字符作为节点，字符能相互走到的关系作为无向边，该题目则为问每一个节点所属的图中节点个数，直接用并查集将存在边的节点合并为同一个集合，同时维护一下集合的节点个数即可。
代码:
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int far[maxn];
int dsu_sz[maxn];
void dsu_init(int n)
{
repd(i, 0, n)
{
far[i] = i;
dsu_sz[i] = 1;
}
}
int findpar(int x)
{
if (x == far[x])
{
return x;
} else
{
return far[x] = findpar(far[x]);
}
}
void mg(int x, int y)
{
x = findpar(x);
y = findpar(y);
if (x == y)
return;
if (dsu_sz[x] > dsu_sz[y])
{
dsu_sz[x] += dsu_sz[y];
far[y] = x;
} else
{
dsu_sz[y] += dsu_sz[x];
far[x] = y;
}
}
int n, m;
int q;
char s[3004][3004];
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readint();
m = readint();
q = readint();
repd(i, 1, n)
{
scanf("%s", s[i] + 1);
}
dsu_init(n * m);
repd(i, 1, n)
{
repd(j, 1, m)
{
if (i + 1 <= n && s[i][j] != s[i + 1][j])
{
mg(m * (i - 1) + j, m * (i) + j);
}
if (j + 1 <= m && s[i][j] != s[i][j + 1])
{
mg(m * (i - 1) + j, m * (i - 1) + j + 1);
}
}
}
int x, y;
while (q--)
{
x = readint();
y = readint();
printf("%d\n", dsu_sz[findpar(m * (x - 1) + y)] );
}
return 0;
}
I-Prime
思路：
线性筛一下质数，然后求个前缀和即可。
代码：
const int N = 10000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int no[N];
int tot;
int sshu[N];
int sum[N];
void prepare()
{
for (int i = 2; i < N; i++)
{
if (!no[i])
sshu[++tot] = i;
for (int j = 1; j <= tot && sshu[j]*i < N; j++)
{
no[sshu[j]*i] = 1;
if (i % sshu[j] == 0)
{
no[sshu[j]*i] = 1;
break;
}
}
}
for (int i = 2; i < N; i++)
{
sum[i] += sum[i - 1];
if (!no[i])
{
sum[i]++;
}
}
}
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
prepare();
int t;
t = readint();
while (t--)
{
int l = readint();
int r = readint();
printf("%d\n", sum[r] - sum[l - 1] );
}
return 0;
}
J-Compare
思路：
用字符串表示数字，比较大小时，先比较长度，长度如果相等再比较字典序即可。
代码：
#define DEBUG_Switch 0
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
string a, b;
cin >> a >> b;
if (sz(a) > sz(b))
{
cout << ">" << endl;
} else if (sz(a) < sz(b))
{
cout << "<" << endl;
} else
{
if (a == b)
{
cout << "=" << endl;
} else if (a > b)
{
cout << ">" << endl;
} else
{
cout << "<" << endl;
}
}
return 0;
}
K-Walk
思路：
从\((1,1)\)走到\((n,m)\)一共有\(n + m - 2\)步，有\(n-1\)步向右走，所以答案为\(C(n + m - 2,n-1)\)
预处理一下阶乘和逆元即可快速求解。
代码:
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
const ll mod = 1e9 + 7;
ll fac[maxn], inv[maxn];
void pre()
{
fac[0] = 1;
for (int i = 1; i < maxn; i++) fac[i] = fac[i - 1] * i % mod;
inv[maxn - 1] = powmod(fac[maxn - 1], mod - 2, mod);
for (int i = maxn - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}
ll C(int a, int b)
{
if (b > a || b < 0) return 0;
return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
pre();
int t;
t = readint();
while (t--)
{
int n = readint();
int m = readint();
printf("%lld\n", C(n + m - 2, n - 1) );
}
return 0;
}
L-Defeat the monster
思路：
排序后二分一下，然后维护答案即可。
代码：
int n;
ll a[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","r",stdin);
n = readint();
repd(i, 1, n)
{
a[i] = readll();
}
sort(a + 1, a + 1 + n);
int ans = 0;
repd(i, 1, n)
{
int pos = upper_bound(a + i, a + 1 + n, a[i] + 5) - a;
ans = max(ans, pos - i);
}
printf("%d\n", ans);
return 0;
}
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