cf670d2 二分贪心

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Problem: cf670d2
Analyse:
经典的二分加贪心模型,我太高估了暴力的复杂度了….
先考虑最坏情况,就是只缺一个物品,而且这个物品缺1e9个,这里面明显需要一个logk的算法,
二分答案,然后贪心地去判断,最后要消耗小于k,贪心就满足,注意这里会爆ll,要提前跳出判断函数.
复杂度nlogk.

二分闭区间的模板化写法:

int l = 0, r = INF;while (l < r) {    int mid = (l + r + 1) / 2;    if (ok) l = mid;    else r = mid - 1;}

Code:

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 四  5/ 5 23:59:56 2016 *File Name*     : jy.cpp***********************1599664856@qq.com**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a));#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define rep(i, a, n) for (int i = a; i < n; i++)#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3fffffff;const int maxn = 1e5 + 13;ll A[maxn], B[maxn], n, k;bool ok(ll t) {    ll ret = 0;    rep (i, 0, n) {        if (B[i] <= A[i] * t) {            ret += A[i] * t - B[i];            if (k < ret) return false;        }    }    return k >= ret;}int main(void){#ifdef LOCAL   freopen("/Users/zhaoyang/in.txt", "r", stdin);    //freopen("/Users/zhaoyang/out.txt", "w", stdout);#endif    //ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);    while (cin >> n >> k) {    ll suma = 0, ans = 0 ;    rep (i, 0, n) cin >> A[i], suma += A[i];    rep (i, 0, n) cin >> B[i];    ll mins = INF;    rep (i, 0, n) {        mins = min(mins, B[i] / A[i]);    }    ans += mins;    rep (i, 0, n) {        B[i] -= mins * A[i];    }    ll r = k, l = 0;    while (l < r) {        ll mid = (r + l + 1) / 2;        if (ok(mid)) {            l = mid;        } else r = mid - 1;    }    cout << ans + l << endl;    }    return 0;}

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