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 l, t and m and you should find the largest number r such that l?≤?rand 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.
The first line of input contains three integers A, B 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.
For each query, print its answer in a single line.
2 1 4 1 5 3 3 3 10 7 10 2 6 4 8
4 -1 8 -1
1 5 2 1 5 10 2 7 4
1 2
英语渣渣看题都看半天
首先给出a,b,n
这是一个由a开头，b为公差，长度无限的等差数列
然后n个询问
输入l,t,m
取这个数列从l开始，m个数，每一次这个数列中所有的数字-1
当首尾变成0的时候，可以向后移，问最后t次之后，最长的0序列的右边界是多少
那么我们首先可以确定，对于这个最终的序列而言，假设右边界为r
那么max(h1,h2,...hr)<=t && h1+h2+...hr<=t*m
对于这两个条件，我们进行二分即可
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <algorithm> using namespace std; #define ls 2*i #define rs 2*i+1 #define up(i,x,y) for(i=x;i<=y;i++) #define down(i,x,y) for(i=x;i>=y;i--) #define mem(a,x) memset(a,x,sizeof(a)) #define w(a) while(a) #define LL long long const double pi = acos(-1.0); #define Len 63 #define mod 19999997 const int INF = 0x3f3f3f3f; LL a,b,n; LL l,t,m; LL cal(LL x) { return a+(x-1)*b; } LL getsum(LL r) { return (cal(r)+cal(l))*(r-l+1)/2; } int main() { LL i,j,k,maxn; w(~scanf("%I64d%I64d%I64d",&a,&b,&n)) { w(n--) { scanf("%I64d%I64d%I64d",&l,&t,&m); if(cal(l)>t) { printf("-1\n"); continue; } LL ll = l,lr = (t-a)/b+1,mid; w(ll<=lr) { LL mid = (ll+lr)/2; if(getsum(mid)<=t*m) ll = mid+1; else lr = mid-1; } printf("%d\n",ll-1); } } return 0; }
codeforces535C:Tavas and Karafs(二分)
原文地址：http://blog.csdn.net/libin56842/article/details/45082747
