Codeforces 821 B. Okabe and Banana Trees
2017-06-26 14:58
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题意:
在一个坐标系内,给定一个函数 y=x−m+b,求一个以原点和函数交点为顶点的矩形,并且矩形内坐标和最大。
算法:模拟
代码:
在一个坐标系内,给定一个函数 y=x−m+b,求一个以原点和函数交点为顶点的矩形,并且矩形内坐标和最大。
算法:模拟
代码:
#include <bits/stdc++.h> /* #include <algorithm> #include <cstring> #include <cmath> #include <cstdio> #include <iostream> #include <map> #include <queue> #include <vector> #include <set> */ using namespace std; typedef long long LL; typedef double DB; typedef unsigned int UI; typedef pair<int, int> PII; const int inf = 0x7f7f7f7f; #define rdi() read<int>() #define rdl() read<LL>() #define rds(a) scanf("%s", a) #define mk(i, j) make_pair(i, j) #define pb push_back #define fi first #define se second #define For(i, j, k) for (int i = j; i <= k; i ++) #define Rep(i, j, k) for (int i = j; i >= k; i --) #define Edge(i, u) for (int i = head[u]; i; i = e[i].nxt) template<typename t> t read() { t x = 0; int f = 1; char c = getchar(); while (c > '9' || c < '0') f = c == '-' ? -1 : 1 , c = getchar(); while (c >= '0' && c <= '9') x = x * 10 + c - 48 , c = getchar(); return x * f; } template<typename t> void write(t x) { if (x < 0){ putchar('-'), write(-x); return; } if (x >= 10) write(x / 10); putchar(x % 10 + 48); } long long ans, t, m, b, x, y; int main() { // freopen(".in", "r", stdin); // freopen(".out", "w", stdout); m = rdi(), b = rdi(); x = -m * -b; for (int i = 0; i <= x; i += m) { y = i / -m + b; t = 0; t = t + 1ll * (i + 1) * (1ll * y * (y + 1)) / 2 + 1ll * (i + 1) * i / 2 * (y + 1); ans = max(t, ans); } write(ans); return 0; }
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