Codeforces 812C Sagheer and Nubian Market[二分]
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题意:给了n个物品,每个物品的价值为ai,你可以选取其中的k个,代价是k个ai的和加上他们的下标乘k的和。
问,在给定最大代价S的条件下,最多能取几个,取的最多的情况下,代价是多少。
分析:对能取几个进行二分,每次判断的时候,将原来的数组对ai+i*mid排序,取前mid个,看是否大于S。
以下是代码:
ll n, s, ans1, ans2;struct lx {ll id, v, k;friend bool operator<(const lx &a, const lx &b) {return a.v + a.id*a.k < b.v + b.id*b.k;}}tmp[MAXN], a[MAXN];bool check(int mid) {for (int i = 1; i <= n; ++i)tmp[i] = a[i], tmp[i].k = mid;sort(tmp + 1, tmp + 1 + n);ll sum = 0;for (int i = 1; i <= mid; ++i) {sum += tmp[i].v + tmp[i].id*tmp[i].k;}ans2 = sum;if (sum > s)return false;else return true;}int main() {scanf("%I64d%I64d", &n, &s);for (int i = 1; i <= n; ++i) {scanf("%I64d", &a[i].v);a[i].id = i;}int l = 0, r = n + 1;while (l < r) {int mid = (l + r) >> 1;if (check(mid)) {l = mid + 1;}else r = mid;}l--; ans1 = l;check(l);printf("%I64d %I64d\n", ans1, ans2);}
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