Gym 101201J Shopping (线段树+取模)
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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; }
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