1500 pts 1500 ~ \text{pts} 1500 pts,可以上 2000 2000 2000,Rank 1400 1400 1400 左右。
给定长 n n n 的数组,在后面插入 k k k 个 0 0 0,求最后 n n n 项。
计算可得,需要输出原数组的后 max ( n − k , 0 ) \max(n - k, 0) max(n−k,0) 项。
需要输出 min ( k , n ) \min(k, n) min(k,n) 个 0 0 0。
主要代码:
signed main(){
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cn(n); cn(k);
int a[n]; repn(i, 0, n, 1) cm(a[i]);
repn(i, min(n, k), n, 1){
cout << a[i] << " ";
}
repn(i, 0, min(n, k), 1) cout << 0 << " ";
return 0;
}
时间 a b ‾ : c d ‾ \overline{ab}:\overline{cd} ab:cd 被称为合法的当且仅当:
给定 a b ‾ \overline{ab} ab 和 c d ‾ \overline{cd} cd 的值,求未来最近的合法时间。
暴力即可。注意一天共有 1440 1440 1440 分钟。
下一分钟时,需要改变小时数。
#include
#define int long long
// FOR templates.
#define rep(i, s, n, k) for(int i = s;i <= n;i += k)
#define repn(i, s, n, k) for(int i = s;i < n;i += k)
#define pre(i, s, n, k) for(int i = s;i >= n;i -= k)
#define pren(i, s, n, k) for(int i = s;i > n;i -= k)
// Abbr for STL.
#define pii pair<int, int>
#define pdd pair<double, double>
#define mpi map<int, int>
#define vc vector<int>
// IO templates, proven very useful.
#define cn(n) int n;cin >> n
#define cm(n) cin >> n
// Abbr for funcs.
#define pb push_back
#define mset memset
// #define files
using namespace std;
const int MAXN = 0x3f3f3f3f3f3f3f3fLL;
const int MOD1 = 1000000007LL;
const int MOD2 = 998244353LL;
int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int gcd(int a, int b) {if(b == 0) return a;return gcd(b, a % b);}
inline int lowbit(int x) {return x & (-x);}
inline int lcm(int a, int b) {return a * b / gcd(a, b);}
signed main(){
#ifdef files
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cn(h); cn(m);
repn(i, 0, 1440, 1){
int a, b, c, d;
a = h / 10, b = h % 10, c = m / 10, d = m % 10;
if(a * 10 + c < 24 && b * 10 + d < 60){
cout << h << " " << m << endl;
return 0;
}
m++; if(m == 60) m = 0, h++;
if(h == 24) h = 0;
}
return 0;
}
/*
* things to check
* 1. int overflow or long long memory need
* 2. recursion/array/binary search/dp/loop bounds
* 3. precision
* 4. special cases(n=1,bounds)
* 5. delete debug statements
* 6. initialize(especially multi-tests)
* 7. = or == , n or m ,++ or -- , i or j , > or >= , < or <=
* 8. keep it simple and stupid
* 9. do not delete, use // instead
* 10. operator priority
* 11. is there anything extra to output?
* 12. THINK TWICE CODE ONCE, THINK ONCE DEBUG FOREVER
* 13. submit ONCE, AC once. submit twice, WA forever
* 14. calm down and you'll get good rank
* 15. even a bit wrong scores zero
* 16. ...
**/
一共 n n n 个用户和 q q q 个操作。操作如下:
1 a b
: a a a 用户关注了 b b b 用户;2 a b
: a a a 用户取关了 b b b 用户;3 a b
:询问 a a a 和 b b b 是否互关。用一个 map
存储关系。
映射可以极好地存储各种复杂的数据。
#include
#define int long long
// FOR templates.
#define rep(i, s, n, k) for(int i = s;i <= n;i += k)
#define repn(i, s, n, k) for(int i = s;i < n;i += k)
#define pre(i, s, n, k) for(int i = s;i >= n;i -= k)
#define pren(i, s, n, k) for(int i = s;i > n;i -= k)
// Abbr for STL.
#define pii pair<int, int>
#define pdd pair<double, double>
#define mpi map<int, int>
#define vc vector<int>
// IO templates, proven very useful.
#define cn(n) int n;cin >> n
#define cm(n) cin >> n
// Abbr for funcs.
#define pb push_back
#define mset memset
// #define files
using namespace std;
const int MAXN = 0x3f3f3f3f3f3f3f3fLL;
const int MOD1 = 1000000007LL;
const int MOD2 = 998244353LL;
int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int gcd(int a, int b) {if(b == 0) return a;return gcd(b, a % b);}
inline int lowbit(int x) {return x & (-x);}
inline int lcm(int a, int b) {return a * b / gcd(a, b);}
map<pii, int> mp;
signed main(){
#ifdef files
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cn(n); cn(q);
repn(i, 0, q, 1){
cn(t); cn(a); cn(b);
if(t == 1) mp[make_pair(a, b)] = 1;
else if(t == 2) mp[make_pair(a, b)] = 0;
else cout << ((mp[make_pair(a, b)] && mp[make_pair(b, a)]) ? "Yes" : "No") << endl;
}
return 0;
}
/*
* things to check
* 1. int overflow or long long memory need
* 2. recursion/array/binary search/dp/loop bounds
* 3. precision
* 4. special cases(n=1,bounds)
* 5. delete debug statements
* 6. initialize(especially multi-tests)
* 7. = or == , n or m ,++ or -- , i or j , > or >= , < or <=
* 8. keep it simple and stupid
* 9. do not delete, use // instead
* 10. operator priority
* 11. is there anything extra to output?
* 12. THINK TWICE CODE ONCE, THINK ONCE DEBUG FOREVER
* 13. submit ONCE, AC once. submit twice, WA forever
* 14. calm down and you'll get good rank
* 15. even a bit wrong scores zero
* 16. ...
**/
给定数组 a a a,做以下操作:
1 x
:给 a a a 的所有点赋值 x x x;2 i x
:给 a i a_i ai 赋值 x x x;3 i
:单点查询。对于每个点建一个向量存储修改值、修改时间和前缀和。
查询时就用这些东西二分 upper_bound
。
#include
#define int long long
// FOR templates.
#define rep(i, s, n, k) for(int i = s;i <= n;i += k)
#define repn(i, s, n, k) for(int i = s;i < n;i += k)
#define pre(i, s, n, k) for(int i = s;i >= n;i -= k)
#define pren(i, s, n, k) for(int i = s;i > n;i -= k)
// Abbr for STL.
#define pii pair<int, int>
#define pdd pair<double, double>
#define mpi map<int, int>
#define vc vector<int>
// IO templates, proven very useful.
#define cn(n) int n;cin >> n
#define cm(n) cin >> n
// Abbr for funcs.
#define pb push_back
#define mset memset
// #define files
using namespace std;
const int MAXN = 0x3f3f3f3f3f3f3f3fLL;
const int MOD1 = 1000000007LL;
const int MOD2 = 998244353LL;
int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int gcd(int a, int b) {if(b == 0) return a;return gcd(b, a % b);}
inline int lowbit(int x) {return x & (-x);}
inline int lcm(int a, int b) {return a * b / gcd(a, b);}
vc vec[200001], vec1[200001], vec2[200001];
signed main(){
#ifdef files
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cn(n); int a[n + 1];
rep(i, 1, n, 1) cm(a[i]);
rep(i, 1, n, 1){
vec[i].pb(0); vec[i].pb(a[i]);
vec1[i].pb(-1); vec1[i].pb(0);
vec2[i].pb(0); vec2[i].pb(a[i]);
}
int lst = -1, nm = 0; cn(q);
rep(i, 1, q, 1){
cn(op); if(op == 1){
cn(x); lst = i, nm = x;
}
else if(op == 2){
cn(k); cn(x);
vec[k].pb(x); vec1[k].pb(i);
vec2[k].pb(vec2[k][vec2[k].size() - 1] + x);
}
else{
cn(k);
int pos = (int)(upper_bound(vec1[k].begin(), vec1[k].end(), lst) - vec1[k].begin());
cout << vec2[k][vec2[k].size() - 1] - vec2[k][pos - 1] + nm << endl;
}
}
return 0;
}
/*
* things to check
* 1. int overflow or long long memory need
* 2. recursion/array/binary search/dp/loop bounds
* 3. precision
* 4. special cases(n=1,bounds)
* 5. delete debug statements
* 6. initialize(especially multi-tests)
* 7. = or == , n or m ,++ or -- , i or j , > or >= , < or <=
* 8. keep it simple and stupid
* 9. do not delete, use // instead
* 10. operator priority
* 11. is there anything extra to output?
* 12. THINK TWICE CODE ONCE, THINK ONCE DEBUG FOREVER
* 13. submit ONCE, AC once. submit twice, WA forever
* 14. calm down and you'll get good rank
* 15. even a bit wrong scores zero
* 16. ...
**/
给出一个矩阵,每个矩阵一个数 a i , j ≤ n a_{i, j} \le n ai,j≤n。
现有一个 h × w h\times w h×w 的纸片,求当这个纸片的右上角在 ( i , j ) (i,j) (i,j) 上时剩下的数的种类。
将每个颜色记录下来,建立 n n n 个不等式方程。
对于每个格子直接暴力求解。
#include
#define int long long
// FOR templates.
#define rep(i, s, n, k) for(int i = s;i <= n;i += k)
#define repn(i, s, n, k) for(int i = s;i < n;i += k)
#define pre(i, s, n, k) for(int i = s;i >= n;i -= k)
#define pren(i, s, n, k) for(int i = s;i > n;i -= k)
// Abbr for STL.
#define pii pair<int, int>
#define pdd pair<double, double>
#define mpi map<int, int>
#define vc vector<int>
// IO templates, proven very useful.
#define cn(n) int n;cin >> n
#define cm(n) cin >> n
// Abbr for funcs.
#define pb push_back
#define mset memset
// #define files
using namespace std;
const int MAXN = 0x3f3f3f3f3f3f3f3fLL;
const int MOD1 = 1000000007LL;
const int MOD2 = 998244353LL;
int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int gcd(int a, int b) {if(b == 0) return a;return gcd(b, a % b);}
inline int lowbit(int x) {return x & (-x);}
inline int lcm(int a, int b) {return a * b / gcd(a, b);}
vc vec[200001], vec1[200001], vec2[200001];
signed main(){
#ifdef files
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cn(n); int a[n + 1];
rep(i, 1, n, 1) cm(a[i]);
rep(i, 1, n, 1){
vec[i].pb(0); vec[i].pb(a[i]);
vec1[i].pb(-1); vec1[i].pb(0);
vec2[i].pb(0); vec2[i].pb(a[i]);
}
int lst = -1, nm = 0; cn(q);
rep(i, 1, q, 1){
cn(op); if(op == 1){
cn(x); lst = i, nm = x;
}
else if(op == 2){
cn(k); cn(x);
vec[k].pb(x); vec1[k].pb(i);
vec2[k].pb(vec2[k][vec2[k].size() - 1] + x);
}
else{
cn(k);
int pos = (int)(upper_bound(vec1[k].begin(), vec1[k].end(), lst) - vec1[k].begin());
cout << vec2[k][vec2[k].size() - 1] - vec2[k][pos - 1] + nm << endl;
}
}
return 0;
}
/*
* things to check
* 1. int overflow or long long memory need
* 2. recursion/array/binary search/dp/loop bounds
* 3. precision
* 4. special cases(n=1,bounds)
* 5. delete debug statements
* 6. initialize(especially multi-tests)
* 7. = or == , n or m ,++ or -- , i or j , > or >= , < or <=
* 8. keep it simple and stupid
* 9. do not delete, use // instead
* 10. operator priority
* 11. is there anything extra to output?
* 12. THINK TWICE CODE ONCE, THINK ONCE DEBUG FOREVER
* 13. submit ONCE, AC once. submit twice, WA forever
* 14. calm down and you'll get good rank
* 15. even a bit wrong scores zero
* 16. ...
**/
状压 dp (?) 咕咕咕