Gym 101201J Shopping (线段树+取模)

题意：给定 n 个物品，然后有 m 个人买东西，他们有 x 元钱，然后从 l - r 这个区间内买东西，对于每个物品都尽可能多的买，问你最少剩下多少钱。

析：对于物品，尽可能多的买的意思就是对这个物品价格取模，但是对于价格比我的钱还多，那么就没有意义，对取模比我的钱少的，那取模至少减少一半，所以最多只要60多次就可以结束，为了快速找到第一个比我的钱少的，使用线段树。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 2e5 + 50;

const LL mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

LL minv[maxn<<2];

LL a[maxn];

void push_up(int rt){ minv[rt] = min(minv[rt<<1], minv[rt<<1|1]); }

void build(int l, int r, int rt){

  if(l == r){ scanf("%I64d", minv + rt);  a[l] = minv[rt]; return ; }

  int m = l + r >> 1;

  build(lson);

  build(rson);

  pu(rt);

}

int query(int L, int R, LL val, int l, int r, int rt){

  if(minv[rt] > val)  return -1;

  if(l == r)  return minv[rt] <= val ? l : -1;

  int m = l + r >> 1;

  int ans = -1;

  if(L <= m){

    if(minv[rt<<1] <= val)  ans = query(L, R, val, lson);

  }

  if(R > m && ans == -1){

    if(minv[rt<<1|1] <= val)  ans = query(L, R, val, rson);

  }

  return ans;

}

int main(){

  scanf("%d %d", &n, &m);

  build(all);

  while(m--){

    LL money;

    int l, r;

    scanf("%I64d %d %d", &money, &l, &r);

    while (money && l <= r){

	  int cur = query(l, r, money, all);

	  if (cur == -1) break;

	  money %= a[cur];

	  l = cur + 1;

    }

    printf("%I64d\n", money);

  }

  return 0;

}