文章标题 codeforce 812C Sagheer and Nubian Market(二分)
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Sagheer and Nubian Market
题目链接
题意:有N个物品(1~n)都有个价值ai,然后S个单位钱,然后要在这n个物品中取k个物品,使得在S 内,k尽可能大,k相等时,所需的钱尽可能少。没件物品的价格是 基础价格ai+k*i(k为所买物品的数目,i为所选物品的下标)。
分析:将所需的物品数二分,每次验证的时候将价格算一下,然后拍一下序。注:注意在计算价格的时候有可能爆int。
代码:
#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;typedef pair<int,int> pii;long long a[100005];long long b[100005];long long T;long long n,s;long long judge(int x){ for (long long i=1;i<=n;i++){ b[i]=a[i]+(i)*x;//计算价格 } sort(b+1,b+n+1); long long sum=0; for (int i=1;i<=x;i++){ sum+=b[i]; } if (sum>s)return 0; T=sum; return 1;}int main (){ while (scanf ("%d%d",&n,&s)!=EOF){ for (int i=1;i<=n;i++)scanf ("%I64d",&a[i]); T=0; long long hi=n,lo=0; long long ans=0; long long mid; long long k=0;//保存所取的件数 while (lo<=hi){//二分 mid=(lo+hi)>>1; if (judge(mid)){ lo=mid+1; k=mid; ans=T; }else { hi=mid-1; } } printf ("%I64d %I64d\n",k,ans); } return 0;}
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