Atcoder评测机没有ONLINE_JUDGE这个宏！！
A,B巨水题
C - Walk on Multiplication Table
Description
给一个数mul，求满足$x*y=mul && min{x+y-2}$
$mul leq 10^{12}$
Solution
由均值不等式可知$x+y geq 2*sqrt{xy}$,给定xy,那么$min{x+y-2}$一定出现在$sqrt{xy}$附近，扫一遍即可
1 #include <algorithm> 2 #include <bits/stdc++.h> 3 #include <cctype> 4 #include <cmath> 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cstring> 8 #include <ctime> 9 #include <iostream> 10 #include <map> 11 #include <numeric> 12 #include <queue> 13 #include <set> 14 #include <stack> 15 #if __cplusplus >= 201103L 16 #include <unordered_map> 17 #include <unordered_set> 18 #endif 19 #include <vector> 20 #define lson rt << 1, l, mid 21 #define rson rt << 1 | 1, mid + 1, r 22 #define LONG_LONG_MAX 9223372036854775807LL 23 #define pblank putchar(' ') 24 #define ll LL 25 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) 26 using namespace std; 27 typedef long long ll; 28 typedef long double ld; 29 typedef unsigned long long ull; 30 typedef pair<int, int> P; 31 ll n, m, k; 32 const int maxn = 1e7 + 10; 33 template <class T> 34 inline T read() 35 { 36 int f = 1; 37 T ret = 0; 38 char ch = getchar(); 39 while (!isdigit(ch)) 40 { 41 if (ch == '-') 42 f = -1; 43 ch = getchar(); 44 } 45 while (isdigit(ch)) 46 { 47 ret = (ret << 1) + (ret << 3) + ch - '0'; 48 ch = getchar(); 49 } 50 ret *= f; 51 return ret; 52 } 53 template <class T> 54 inline void write(T n) 55 { 56 if (n < 0) 57 { 58 putchar('-'); 59 n = -n; 60 } 61 if (n >= 10) 62 { 63 write(n / 10); 64 } 65 putchar(n % 10 + '0'); 66 } 67 template <class T> 68 inline void writeln(const T &n) 69 { 70 write(n); 71 puts(""); 72 } 73 int main(int argc, char const *argv[]) 74 { 75 #ifndef ONLINE_JUDGE 76 // freopen("in.txt", "r", stdin); 77 // freopen("out.txt", "w", stdout); 78 #endif 79 n = read<ll>(); 80 ll up = (ll)sqrt(n * 1.0); 81 ll res = n - 1; 82 for (ll i = up; i >= 1; i--) 83 if (n % i == 0) 84 { 85 res = i + n / i - 2; 86 break; 87 } 88 writeln(res); 89 return 0; 90 }
D - Water
Description
给一个底部为正方形边长为a，高为b的无盖水瓶，以及x单位体积的水，问最大倾斜多少角度可以使水刚好不溢出。
Solution
赛场降智。只想到第一种情况，测试样例都过不完心态崩了。
贴一张nb网友做的图。
分两种情况讨论就行了，注意转换为角度值。
1 #include <algorithm> 2 #include <cctype> 3 #include <cmath> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <iomanip> 8 #include <iostream> 9 #include <map> 10 #include <numeric> 11 #include <queue> 12 #include <set> 13 #include <stack> 14 #if __cplusplus >= 201103L 15 #include <unordered_map> 16 #include <unordered_set> 17 #endif 18 #include <vector> 19 #define lson rt << 1, l, mid 20 #define rson rt << 1 | 1, mid + 1, r 21 #define LONG_LONG_MAX 9223372036854775807LL 22 #define pblank putchar(' ') 23 #define ll LL 24 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) 25 using namespace std; 26 typedef long long ll; 27 typedef long double ld; 28 typedef unsigned long long ull; 29 typedef pair<int, int> P; 30 int n, m, k; 31 const int maxn = 1e5 + 10; 32 template <class T> 33 inline T read() 34 { 35 int f = 1; 36 T ret = 0; 37 char ch = getchar(); 38 while (!isdigit(ch)) 39 { 40 if (ch == '-') 41 f = -1; 42 ch = getchar(); 43 } 44 while (isdigit(ch)) 45 { 46 ret = (ret << 1) + (ret << 3) + ch - '0'; 47 ch = getchar(); 48 } 49 ret *= f; 50 return ret; 51 } 52 template <class T> 53 inline void write(T n) 54 { 55 if (n < 0) 56 { 57 putchar('-'); 58 n = -n; 59 } 60 if (n >= 10) 61 { 62 write(n / 10); 63 } 64 putchar(n % 10 + '0'); 65 } 66 template <class T> 67 inline void writeln(const T &n) 68 { 69 write(n); 70 puts(""); 71 } 72 const ld PI = acos(-1); 73 ld solve(int a, int b, int x) 74 { 75 if (2.0 * x >= a * a * b) 76 return (ld)atan((2 * a * a * b * 1.0 - 2 * x) / (a * a * a * 1.0)); 77 return PI / 2 - atan(2 * x * 1.0 / (a * b * b)); 78 } 79 int main(int argc, char const *argv[]) 80 { 81 #ifndef ONLINE_JUDGE 82 // freopen("in.txt", "r", stdin); 83 // freopen("out.txt", "w", stdout); 84 #endif 85 86 int a, b, x; 87 cin >> a >> b >> x; 88 cout << fixed << setprecision(7) << solve(a, b, x) * 180.0 / PI; 89 return 0; 90 }
E - Gluttony
Description
给出两个长为n的序列A,F以及一个操作数上界k(可以不将操作次数用完)，每次操作可以使A序列里的任一个值减一。
操作结束后，将A,F元素两两配对,求配对的最小乘积。
Solution
最近总是碰见这种题。
要求最小乘积，我们肯定得把最小的A和最大的F配对。
对于答案最小乘积而言，如果我们可以得到答案为x，那么一定可以得到一个大于x的答案（减少操作数或者将配对情况交换）
也就是说决策具有单调性。
那么就很显然可以二分来搞。
每次check时将配对乘积大于答案的减去最小的次数，判断最后次数是否小于等于k
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; 30 ll k; 31 const int maxn = 2e5 + 10; 32 template <class T> 33 inline T read() 34 { 35 int f = 1; 36 T ret = 0; 37 char ch = getchar(); 38 while (!isdigit(ch)) 39 { 40 if (ch == '-') 41 f = -1; 42 ch = getchar(); 43 } 44 while (isdigit(ch)) 45 { 46 ret = (ret << 1) + (ret << 3) + ch - '0'; 47 ch = getchar(); 48 } 49 ret *= f; 50 return ret; 51 } 52 template <class T> 53 inline void write(T n) 54 { 55 if (n < 0) 56 { 57 putchar('-'); 58 n = -n; 59 } 60 if (n >= 10) 61 { 62 write(n / 10); 63 } 64 putchar(n % 10 + '0'); 65 } 66 template <class T> 67 inline void writeln(const T &n) 68 { 69 write(n); 70 puts(""); 71 } 72 ll a[maxn], f[maxn]; 73 inline int judge(ll x) 74 { 75 ll cnt = 0; 76 for (int i = 1; i <= n; i++) 77 if (a[i] * f[n - i + 1] > x) 78 { 79 ll tmp = a[i] * f[n - i + 1] - x; 80 cnt += tmp / f[n - i + 1]; 81 if (tmp % f[n - i + 1]) 82 ++cnt; 83 } 84 return cnt <= k; 85 } 86 int main(int argc, char const *argv[]) 87 { 88 #ifndef ONLINE_JUDGE 89 // freopen("in.txt", "r", stdin); 90 // freopen("out.txt", "w", stdout); 91 #endif 92 n = read<int>(); 93 k = read<ll>(); 94 for (int i = 1; i <= n; i++) 95 a[i] = read<ll>(); 96 for (int i = 1; i <= n; i++) 97 f[i] = read<ll>(); 98 sort(a + 1, a + 1 + n); 99 sort(f + 1, f + 1 + n); 100 ll l = 0, r = 1e12 + 1; 101 ll res = 0; 102 while (l <= r) 103 { 104 ll mid = l + r >> 1; 105 if (judge(mid)) 106 { 107 res = mid; 108 r = mid - 1; 109 } 110 else 111 l = mid + 1; 112 } 113 writeln(res); 114 return 0; 115 }