CodeForces 449AJzzhu and Chocolate(贪心)

来源：互联网发布：大数据的定义是什么编辑：程序博客网时间：2024/05/16 19:21

CodeForces 449A

题目大意：
一块N*M的巧克力，指定K次切割，切割的要求看题目。问能否使得最小的那块巧克力的面积最大。如果不能进行K次切割输出-1.

解题思路：
要使得切割后最小的巧克力的面积最大，那么尽量是朝一个方向切割是最优的。

代码：

#include <cstdio>typedef long long ll;ll N, M, K;ll max (ll a, ll b) {    return a > b ? a: b;}ll min(ll a, ll b) {    return a < b ? a: b;}ll handle(ll x, ll y) {    ll a = N/(x + 1);    a = min(a, N - a*x);    ll b = M/(y + 1);    b = min(b, M - b*y);    return a * b;}ll solve () {    ll x = max(K - N + 1, 0);    ll y = max(K - M + 1, 0);    return max(handle(K - x, x), handle(y, K - y));}int main () {    while (scanf ("%lld%lld%lld", &N, &M, &K) != EOF) {        if (K > (N + M) - 2) {            printf ("-1\n");            continue;        }/*      if (N > M) {            ll t = N;            N = M;            M = t;        }        if (K + 1 <= M) {            if (N % (K + 1) == 0)                 printf ("%lld\n", N /(K + 1) * M);            else if (M % (K + 1) == 0)                 printf ("%lld\n", M /(K + 1) * N);            else                 printf ("%lld\n", solve());        } else             printf ("%lld\n", max(min(N/(K - M + 2), N - (N/(K - M + 2))*(K - M + 1)), min(M/(K - N + 2), M - (M/(K - N + 2) * (K - N + 1)))));*/        printf ("%lld\n", solve());    }}

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