A. Prefixes
Description
给出一个只包含a,b的偶数长度字符串。
每次可以将序列上某一个值a->b或者b->a。
问最小的操作次数满足从起始位置开始偶数长度的子串中a，b个数相等。
Solution
模拟。
B. Shooting
Description
Solution
按权值排序一遍，贪心。
C. White Sheet
Description
给出矩形A，B，C的左下和右上坐标。
求A是否完全被B,C覆盖。
Solution
求A与B的交集AB，A与C的交集AC，ABC的交集ABC。
容斥一下如果A=AB+AC-ABC则证明完全被覆盖。
判断两个矩形是否相交
$$max(A_{lx},B_{lx}) leq min(A_{rx},B_{rx}) && max(A_{ly},B_{ly}) leq min(A_{ry},B_{ry})$$
交点坐标
$$lx=max(A_{lx},B_{lx}),ly=max(A_{ly},B_{ly}),rx=min(A_{rx},B_{rx}),ry= min(A_{ry},B_{ry})$$
1 #include <algorithm> 2 #include <cctype> 3 #include <cmath> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <iostream> 8 #include <map> 9 #include <numeric> 10 #include <queue> 11 #include <set> 12 #include <stack> 13 #if __cplusplus >= 201103L 14 #include <unordered_map> 15 #include <unordered_set> 16 #endif 17 #include <vector> 18 #define lson rt << 1, l, mid 19 #define rson rt << 1 | 1, mid + 1, r 20 #define LONG_LONG_MAX 9223372036854775807LL 21 #define pblank putchar(' ') 22 #define ll LL 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) 24 using namespace std; 25 typedef long long ll; 26 typedef long double ld; 27 typedef unsigned long long ull; 28 typedef pair<int, int> P; 29 int n, m, k; 30 const int maxn = 1e5 + 10; 31 template <class T> 32 inline T read() 33 { 34 int f = 1; 35 T ret = 0; 36 char ch = getchar(); 37 while (!isdigit(ch)) 38 { 39 if (ch == '-') 40 f = -1; 41 ch = getchar(); 42 } 43 while (isdigit(ch)) 44 { 45 ret = (ret << 1) + (ret << 3) + ch - '0'; 46 ch = getchar(); 47 } 48 ret *= f; 49 return ret; 50 } 51 template <class T> 52 inline void write(T n) 53 { 54 if (n < 0) 55 { 56 putchar('-'); 57 n = -n; 58 } 59 if (n >= 10) 60 { 61 write(n / 10); 62 } 63 putchar(n % 10 + '0'); 64 } 65 template <class T> 66 inline void writeln(const T &n) 67 { 68 write(n); 69 puts(""); 70 } 71 template <typename T> 72 void _write(const T &t) 73 { 74 write(t); 75 } 76 template <typename T, typename... Args> 77 void _write(const T &t, Args... args) 78 { 79 write(t), pblank; 80 _write(args...); 81 } 82 template <typename T, typename... Args> 83 inline void write_line(const T &t, const Args &... data) 84 { 85 _write(t, data...); 86 } 87 struct node 88 { 89 int x1, y1, x2, y2; 90 node() {} 91 node(int a, int b, int c, int d) 92 { 93 x1 = a, y1 = b, x2 = c, y2 = d; 94 } 95 ll area() 96 { 97 return (x2 - x1) * 1LL * (y2 - y1); 98 } 99 }; 100 ll cal(const node &t1, const node &t2) 101 { 102 node tmp; 103 tmp.y2 = min(t2.y2, t1.y2); 104 tmp.y1 = max(t1.y1, t2.y1); 105 tmp.x2 = min(t1.x2, t2.x2); 106 tmp.x1 = max(t1.x1, t2.x1); 107 return tmp.area(); 108 } 109 inline int check(const node &t1, const node &t2) 110 { 111 node tmp; 112 tmp.x1 = max(t1.x1, t2.x1); 113 tmp.x2 = min(t1.x2, t2.x2); 114 tmp.y1 = max(t1.y1, t2.y1); 115 tmp.y2 = min(t1.y2, t2.y2); 116 if (tmp.x1 > tmp.x2 || tmp.y1 > tmp.y2) 117 return 0; 118 return 1; 119 } 120 inline node intersection(const node &t1, const node &t2) 121 { 122 node tmp; 123 tmp.x1 = max(t1.x1, t2.x1); 124 tmp.x2 = min(t1.x2, t2.x2); 125 tmp.y1 = max(t1.y1, t2.y1); 126 tmp.y2 = min(t1.y2, t2.y2); 127 return tmp; 128 } 129 int main(int argc, char const *argv[]) 130 { 131 #ifndef ONLINE_JUDGE 132 freopen("in.txt", "r", stdin); 133 // freopen("out.txt","w", stdout); 134 #endif 135 vector<node> vec(3); 136 for (int i = 0; i < 3; i++) 137 { 138 int x1 = read<int>(), y1 = read<int>(), x2 = read<int>(), y2 = read<int>(); 139 vec[i] = node(x1, y1, x2, y2); 140 } 141 ll p1 = 0, p2 = 0; 142 if (check(vec[0],vec[1])) 143 p1 = intersection(vec[0], vec[1]).area(); 144 if (check(vec[0],vec[2])) 145 p2 = intersection(vec[0], vec[2]).area(); 146 ll now = p1 + p2; 147 if (check(vec[1], vec[2])) 148 if (check(vec[0],intersection(vec[1],vec[2]))) 149 now -= intersection(vec[0], intersection(vec[1], vec[2])).area(); 150 if (now >= vec[0].area()) 151 puts("NO"); 152 else 153 puts("YES"); 154 return 0; 155 }
D. Swords
Description
Solution
读题找规律。
找出最大的数mx，然后求所有数与mx的差的和以及gcd。
最后答案为sum/gcd,gcd。
1 #include <algorithm> 2 #include <numeric> 3 #include <cctype> 4 #include <cmath> 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cstring> 8 #include <iostream> 9 #include <map> 10 #include <queue> 11 #include <set> 12 #include <stack> 13 #if __cplusplus >= 201103L 14 #include <unordered_map> 15 #include <unordered_set> 16 #endif 17 #include <vector> 18 #define lson rt << 1, l, mid 19 #define rson rt << 1 | 1, mid + 1, r 20 #define LONG_LONG_MAX 9223372036854775807LL 21 #define pblank putchar(' ') 22 #define ll LL 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) 24 using namespace std; 25 typedef long long ll; 26 typedef long double ld; 27 typedef unsigned long long ull; 28 typedef pair<int, int> P; 29 int n, m, k; 30 const int maxn = 2e5 + 10; 31 template <class T> 32 inline T read() 33 { 34 int f = 1; 35 T ret = 0; 36 char ch = getchar(); 37 while (!isdigit(ch)) 38 { 39 if (ch == '-') 40 f = -1; 41 ch = getchar(); 42 } 43 while (isdigit(ch)) 44 { 45 ret = (ret << 1) + (ret << 3) + ch - '0'; 46 ch = getchar(); 47 } 48 ret *= f; 49 return ret; 50 } 51 template <class T> 52 inline void write(T n) 53 { 54 if (n < 0) 55 { 56 putchar('-'); 57 n = -n; 58 } 59 if (n >= 10) 60 { 61 write(n / 10); 62 } 63 putchar(n % 10 + '0'); 64 } 65 template <class T> 66 inline void writeln(const T &n) 67 { 68 write(n); 69 puts(""); 70 } 71 template <typename T> 72 void _write(const T &t) 73 { 74 write(t); 75 } 76 template <typename T, typename... Args> 77 void _write(const T &t, Args... args) 78 { 79 write(t), pblank; 80 _write(args...); 81 } 82 template <typename T, typename... Args> 83 inline void write_line(const T &t, const Args &... data) 84 { 85 _write(t, data...); 86 } 87 int a[maxn]; 88 int main(int argc, char const *argv[]) 89 { 90 #ifndef ONLINE_JUDGE 91 freopen("in.txt","r", stdin); 92 // freopen("out.txt","w", stdout); 93 #endif 94 n = read<int>(); 95 int mx = 0; 96 for (int i = 1; i <= n;i++) 97 a[i] = read<int>(), mx = max(a[i], mx); 98 int gcd = 0; 99 ll sum = 0; 100 for (int i = 1; i <= n;i++) 101 if (a[i]==mx) 102 continue; 103 else 104 sum += mx - a[i], gcd = __gcd(mx - a[i], gcd); 105 write_line(sum/gcd, gcd); 106 return 0; 107 }
E. Numerical Sequence
Description
给出序列112123123412345....
给q次查询，每次查询序列内的第k个数字（仅包含0-9）
Solution
只想到了简单版本的做法。
简单版本k的数量级是1e9，可以在根号复杂度下模拟完成求解。
复杂版本的题解：
长度为len的数的个数为$10^{len}-10^{len-1}$，那么对于$10^{len-1}-10^{len-1}$这些数的长度都是len。
我们可以记录一个add值，表示上一个数量级的前缀长度。
如1-9范围内，这个前缀长度为9，那么在求解10-99时，这个前缀长度对答案都有贡献，即每一个大于9的数里肯定包含了1-9的前缀
通过前缀累加我们可以很快求得以x结尾的最短序列的长度。即最早出现x的序列长度。
我们可以根据这个来二分一个长度最小但大于k的x序列。
那么我们把{x-1}产生的贡献减去后，就得到了k在单独x序列(只包含12345-...x)里的位置信息。同样二分这个位置，就能得到最后答案。
1 #include <algorithm> 2 #include <numeric> 3 #include <cctype> 4 #include <cmath> 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cstring> 8 #include <iostream> 9 #include <map> 10 #include <queue> 11 #include <set> 12 #include <stack> 13 #if __cplusplus >= 201103L 14 #include <unordered_map> 15 #include <unordered_set> 16 #endif 17 #include <vector> 18 #define lson rt << 1, l, mid 19 #define rson rt << 1 | 1, mid + 1, r 20 #define LONG_LONG_MAX 9223372036854775807LL 21 #define pblank putchar(' ') 22 #define ll LL 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) 24 using namespace std; 25 typedef long long ll; 26 typedef long double ld; 27 typedef unsigned long long ull; 28 typedef pair<int, int> P; 29 int n, m, k; 30 const int maxn = 1e5 + 10; 31 template <class T> 32 inline T read() 33 { 34 int f = 1; 35 T ret = 0; 36 char ch = getchar(); 37 while (!isdigit(ch)) 38 { 39 if (ch == '-') 40 f = -1; 41 ch = getchar(); 42 } 43 while (isdigit(ch)) 44 { 45 ret = (ret << 1) + (ret << 3) + ch - '0'; 46 ch = getchar(); 47 } 48 ret *= f; 49 return ret; 50 } 51 template <class T> 52 inline void write(T n) 53 { 54 if (n < 0) 55 { 56 putchar('-'); 57 n = -n; 58 } 59 if (n >= 10) 60 { 61 write(n / 10); 62 } 63 putchar(n % 10 + '0'); 64 } 65 template <class T> 66 inline void writeln(const T &n) 67 { 68 write(n); 69 puts(""); 70 } 71 template <typename T> 72 void _write(const T &t) 73 { 74 write(t); 75 } 76 template <typename T, typename... Args> 77 void _write(const T &t, Args... args) 78 { 79 write(t), pblank; 80 _write(args...); 81 } 82 template <typename T, typename... Args> 83 inline void write_line(const T &t, const Args &... data) 84 { 85 _write(t, data...); 86 } 87 ll lten[20]; 88 void init() 89 { 90 ll p = 1; 91 for (int i = 0; i < 19;i++) 92 lten[i] = p, p *= 10; 93 } 94 inline ll num(ll x){ 95 return (x + 1) * x / 2; 96 } 97 ll solve(ll x,int op){ 98 ll sum = 0, pre = 1, add = 0; 99 for (int len = 1;;len++){ 100 if (10LL*pre-1<x){ 101 ll cnt = 10 * pre - pre; 102 if (op) 103 { 104 sum += add * cnt + num(cnt) * len; 105 add += cnt * len; 106 } 107 else 108 sum += cnt * len; 109 } 110 else{ 111 ll cnt = x - pre + 1; 112 if (op) 113 sum += add * cnt + num(cnt) * len; 114 else 115 sum += cnt * len; 116 break; 117 } 118 pre *= 10; 119 } 120 return sum; 121 } 122 string tostring(ll x) 123 { 124 string str = ""; 125 while (x) 126 str += char(x % 10 + '0'), x /= 10; 127 reverse(str.begin(), str.end()); 128 return str; 129 } 130 int main(int argc, char const *argv[]) 131 { 132 #ifndef ONLINE_JUDGE 133 freopen("in.txt","r", stdin); 134 // freopen("out.txt","w", stdout); 135 #endif 136 fastIO; 137 int t; 138 cin >> t; 139 while (t--) 140 { 141 ll n; 142 cin >> n; 143 --n; 144 ll l = 1, r = 1e9,res=-1; 145 while(l<=r){ 146 ll mid = l + r >> 1; 147 if (solve(mid,1)>n){ 148 res = mid; 149 r = mid - 1; 150 } 151 else 152 l = mid + 1; 153 } 154 n -= solve(res - 1, 1); 155 156 l = 1, r = res,res=-1; 157 while(l<=r){ 158 ll mid = l + r >> 1; 159 if (solve(mid,0)>n) 160 res = mid, r = mid - 1; 161 else 162 l = mid + 1; 163 } 164 n -= solve(res - 1, 0); 165 cout << tostring(res)[n] << " "; 166 } 167 return 0; 168 }