Codeforces 670d1-magic Powder-1

The title of the topic is to give the amount of material needed to produce a single food, and then give the total quantity of each material, and ask how many such foods can be produced at most.

Greedy. First find out how many of each material is the total number of such materials, and then from small to large sort, and then use an array to save the latter a large amount of material minus the front all the small amount of difference, because for example, there are 3 kinds of materials, each material quantity is 1, 2, 4, need sum[1] = 1, sum[2] = 3, then the greedy calculation.

#include <bits/stdc++.h>using namespace Std;const int MAX = 1e3 + 5;//int A[max], B[max], c[max];struct node{int A, b, C;} Node[max];int Sum[max];bool Comp (Node A, Node B) {return A.C < B.C;} const INT INF = 0x3f3f3f;int Main () {int n, k;scanf ("%d%d", &n, &k); int tot = 0;for (int i = 0; i < n; ++i) {scan F ("%d", &node[i].a);//tot + = node[i].a;} for (int i = 0; i < n; ++i) {scanf ("%d", &node[i].b);//k + = node[i].b;} int smallest = inf;for (int i = 0; i < n; ++i) {node[i].c = node[i].b/node[i].a;//if (Smallest > C[i])//smallest = C[i];} Sort (node, node + N, comp); Sum[0] = node[0].a;for (int i = 1; i < n; ++i) {sum[i] = sum[i-1] + node[i].a;} for (int i = 0; i < n; ++i)//cout << node[i].c << Endl;int i = 0; int res = Node[0].c;while (i <= n-1 && k > 0) {//cout << k << endl;k + = (node[i].b% node[i].a) ;//cout << "ss" << K << endl;if (k >= (sum[i]* (node[i + 1].c-node[i].c)) && node[i + 1].c! = 0 && node[i + 1].c > node[i].c) {k-= (sum[i] * (Node[i + 1].c-node[i].c)), res + = Node[i + 1].c- Node[i].c;//cout << "k =" << k << Endl;} else if (K < (sum[i]* (node[i + 1].c-node[i].c)) && node[i + 1].c! = 0 && node[i + 1].c > Node[i]. c) {//if (node[i + 1].c > node[i].c)//{res + = k/(Sum[i]), Break;//}}else if (node[i + 1].a = = 0) {res + = k/(Sum[i]);// cout << "Here" << Endl;break;} Else//break;i++;//cout << "res" << res << Endl;} cout << Res << Endl;return 0;}

Codeforces 670d1-magic Powder-1