CodeForces 670D2 Magic Powder - 2（二分+贪心）

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http://codeforces.com/contest/670/problem/D2

简单的二分，二分所有可以做的饼干数，然后遍历就可以啦

#include <iostream>#include <stdio.h>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>using namespace std;typedef long long LL;#define N 210000#define INF 0x3f3f3f3f#define met(a, b) memset (a, b, sizeof(a))struct node{    LL one, sum;///做一个饼干需要的材料和已经有的材料    LL num, need;///当前材料可以做的饼干数和做剩下的材料}p[N];bool cmp (node a, node b){    return a.num < b.num;}LL n, k, L, R;LL Search (){    LL l = L, r = R, maxn = 0;    while (l<=r)    {        LL mid = (l+r)/2, S = 0;        for (int i=0; i<n; i++)        {            if (p[i].num < mid)            {                S += p[i].one-p[i].need;                if (mid>p[i].num)                    S += (mid-1-p[i].num)*p[i].one;                if (S > k) break;///第一次没加上这个判断条件WA在第128组测试数据上了            }        }        if (S <= k)        {            maxn = max (maxn, mid);            l = mid+1;        }        else r = mid-1;    }    return maxn;}void Solve (){    sort (p, p+n, cmp);    L = R = p[0].num;    for (int i=0; i<n; i++)        R = max (R, (k+p[i].need)/p[i].one+p[i].num);   LL ans = Search();   printf ("%I64d\n", ans);}int main (){    while (scanf ("%I64d %I64d", &n, &k) != EOF)    {        met (p, 0);        for (int i=0; i<n; i++)            scanf ("%I64d", &p[i].one);        for (int i=0; i<n; i++)        {            scanf ("%I64d", &p[i].sum);            p[i].num = p[i].sum / p[i].one;            p[i].need = p[i].sum % p[i].one;        }        Solve ();    }    return 0;}

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