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

簡單的二分，二分所有可以做的餅干數，然后遍歷就可以啦

#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;///做一個餅干需要的材料和已經有的材料LL num, need;///當前材料可以做的餅干數和做剩下的材料 }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;///第一次沒加上這個判斷條件WA在第128組測試數據上了}}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; }

總結

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