• Codeforces 535C Tavas and Karafs


    题目链接:CF - 535C

    Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.

    Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is si = A + (i - 1) × B.

    For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.

    Now SaDDas asks you n queries. In each query he gives you numbers lt and m and you should find the largest number r such that l ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.

    Input

    The first line of input contains three integers AB and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).

    Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.

    Output

    For each query, print its answer in a single line.

    题目描述:给出一个等差数列,操作严格要求从最左边不为零的连续m个数减去1,最多执行t次后问离最左边最远的位置在哪里。

    算法分析:我们可以二分位置,仔细想想这两个条件:max(hl,hl+1,hl+2,,,,hr)<=t && hl+(hl+1)+(hl+2),,,,hr<=m*t

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<cmath>
     6 #include<algorithm>
     7 #define inf 0x7fffffff
     8 using namespace std;
     9 typedef long long LL;
    10 const LL maxn=1000000+10;
    11 
    12 LL A,B,n;
    13 LL l,t,m;
    14 
    15 LL getValue(LL u) {return A+(u-1)*B; }
    16 LL getSeg(LL u,LL v)
    17 {
    18     return (getValue(u)+getValue(v))*(v-u+1)/2;
    19 }
    20 
    21 int main()
    22 {
    23     while (scanf("%I64d%I64d%I64d",&A,&B,&n)!=EOF)
    24     {
    25         while (n--)
    26         {
    27             scanf("%I64d%I64d%I64d",&l,&t,&m);
    28             LL left=A+(l-1)*B;
    29             if (left>t) printf("-1
    ");
    30             else
    31             {
    32                 LL L=l,R=(t-A)/B+1;
    33                 while (L<=R)
    34                 {
    35                     LL mid=(L+R)>>1;
    36                     if (getSeg(l,mid)<=m*t) L=mid+1;
    37                     else R=mid-1;
    38                 }
    39                 printf("%I64d
    ",L-1);
    40             }
    41         }
    42     }
    43     return 0;
    44 }
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  • 原文地址:https://www.cnblogs.com/huangxf/p/4430837.html
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