A Star
#include <cstdio>
using namespace std;
int n;
int main()
{
scanf("%d", &n);
printf("%d
", 100 - n % 100);
return 0;
}
B uNrEaDaBlE sTrInG
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cctype>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
string str;
bool check()
{
for(int i = 0; i < (int)str.size(); ++ i)
if(i % 2 && islower(str[i])) return 0;
else if(i % 2 == 0 && isupper(str[i])) return 0;
return 1;
}
int main()
{
IOS;
cin >> str;
if(check()) puts("Yes");
else puts("No");
return 0;
}
C Kaprekar Number
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
LL n, k;
LL f(LL x)
{
vector<int> g1, g2;
while(x) g1.push_back(x % 10), g2.push_back(x % 10), x /= 10;
sort(g1.begin(), g1.end(), greater<int>());
sort(g2.begin(), g2.end());
LL res = 0;
for(int i = 0; i < (int)g1.size(); ++ i)
res = res * 10 + g1[i] - g2[i];
return res;
}
int main()
{
IOS;
cin >> n >> k;
LL a0 = n;
for(int i = 1; i <= k; ++ i) a0 = f(a0);
cout << a0 << endl;
return 0;
}
D Base n
特判 (X) 只有一位的情况，无论基底是几，得到的数都是原数，判断原数和 (M) 的大小关系即可
对于其他情况，当基底越大，得到的值也就越大，具有单调性，可以二分基底
计算以 (mid) 为基底的数与 (M) 的大小关系时，直接计算答案会爆 (long ; long), 注意在过程中判断
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
LL m;
string s;
int find()
{
int res = 0;
for(auto x : s) res = max(res, x - '0');
return res;
}
bool check(LL mid)
{
LL res = 0;
for(auto x : s)
{
if(res > m / mid) return 0;
res = res * mid + x - '0';
if(res > m) return 0;
}
return 1;
}
int main()
{
IOS;
cin >> s >> m;
int d = find();
if(s.size() == 1)
{
if((LL)d <= m) cout << "1" << endl;
else cout << "0" << endl;
return 0;
}
LL l = d + 1, res = 0, r = m;
while(l <= r)
{
LL mid = (l + r) >> 1;
if(check(mid))
{
res = mid;
l = mid + 1;
}
else r = mid - 1;
}
if(res) cout << res - d << endl;
else cout << "0" << endl;
return 0;
}
E Train
从 (u) 到达 (v) 的时间为 (leftlceil dfrac{d[u]}{K_i} ight ceil * K_i + T_i)， 建双向边跑 (dijkstra) 即可
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
const int M = 2e5 + 10;
const int N = 1e5 + 10;
const LL INF = 1e18;
int n, m, S, T;
LL d[N];
struct Edge
{
int to, nxt, T, K;
}line[M];
int fist[N], idx;
bool st[N];
struct zt
{
int from;
LL d;
};
bool operator < (zt a, zt b)
{
return a.d > b.d;
}
void add(int x, int y, int z, int w)
{
line[idx] = {y, fist[x], z, w};
fist[x] = idx ++;
}
void heap_dijkstra()
{
for(int i = 1; i <= n; ++ i) d[i] = INF;
priority_queue<zt> q;
q.push({S, 0}), d[S] = 0;
while(!q.empty())
{
zt u = q.top(); q.pop();
if(st[u.from]) continue;
st[u.from] = 1;
for(int i = fist[u.from]; ~i; i = line[i].nxt)
{
int v = line[i].to;
LL tim = (d[u.from] + line[i].K - 1) / line[i].K * line[i].K + line[i].T;
if(d[v] > tim)
{
d[v] = tim;
q.push({v, d[v]});
}
}
}
}
int main()
{
scanf("%d%d%d%d", &n, &m, &S, &T);
memset(fist, -1, sizeof fist);
for(int i = 1; i <= m; ++ i)
{
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
add(a, b, c, d);
add(b, a, c, d);
}
heap_dijkstra();
if(d[T] < INF) printf("%lld
", d[T]);
else puts("-1");
return 0;
}
F Potion
令 (sum A_i) 为从 (n) 个元素中选出 (p) 个元素的权值和
目标是使在满足(p | (X - sum A_i)) 条件下最小化 (dfrac{X - sum A_i}{p})
故对于每个 (p) ，求出在 (sum A_i) mod (p) == (X) mod (p) 的条件下 (sum A_i) 的最大值即可
问题转化为求 (i) 个数中取出 (j) 个且和的余数为 (k) 的最大和
令(f[i][j][k]) 为前 (i) 个数中选择 (j) 个且和的余数为 (k) 的最大和
(f[i][j][k] = max(f[i - 1][j][k], f[i - 1][j - 1][(k - a_i) \% p] + a_i))
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long LL;
const int N = 100 + 20;
const LL INF = 1e18;
LL f[N][N][N], m;
int n, a[N];
int main()
{
IOS;
cin >> n >> m;
for(int i = 1; i <= n; ++ i) cin >> a[i];
LL ans = INF;
for(int p = 1; p <= n; ++ p)
{
memset(f, -0x3f, sizeof f);
f[0][0][0] = 0;
for(int i = 1; i <= n; ++ i)
{
f[i][0][0] = 0;
for(int j = 1; j <= p; ++ j)
for(int k = 0; k < p; ++ k)
f[i][j][k] = max(f[i - 1][j][k], f[i - 1][j - 1][((k - a[i]) % p + p) % p] + a[i]);
}
if(f[n][p][m % p] >= 0) ans = min(ans, (m - f[n][p][m % p]) / p);
}
cout << ans << endl;
return 0;
}