题目来源: 2016-2017 ACM-ICPC Pacific Northwest Regional Contest
J.Shopping
每次找到区间内最左边的小于x的数,然后%一下它,重复以上操作就行了。所以只需要实现一个区间询问最左边的小于x的值就可以了。可以证明每次操作最多log次
解法1:分块。块外暴力,块内提前排好序二分。写挫了莫名t。
解法2:线段树。维护一下区间最小值,显然如果左边的最小值小于等于x那就朝左边递归,否则右边,便可以完成这个操作。
解法3:st表。又不用修改。st表干掉一个log
code:
解法1:挫了不贴了。
解法2:
#include <cstdio> #include <algorithm> #define pb(x) push_back(x) #define rg register const int maxn = 2e5 + 100; typedef unsigned long long ll; inline int readint(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } inline ll readll(){ char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } inline void write(ll x){ if(x>=10LL) write(x/10LL); putchar('0'+x%10LL); } using namespace std; int n, q; ll a[maxn]; struct seg{int l, r; ll x;} tr[maxn << 2]; inline void build(int p, int l, int r) { tr[p].l = l; tr[p].r = r; if(l == r){ tr[p].x = a[l]; return; } int mid = (l + r) >> 1; build(p<<1, l, mid); build(p<<1|1, mid+1, r); tr[p].x = min(tr[p<<1].x, tr[p<<1|1].x); } inline ll ask_mn(int p, int l, int r) { if(tr[p].l == l && tr[p].r == r) return tr[p].x; int mid = (tr[p].l + tr[p].r) >> 1; if(r <= mid) return ask_mn(p<<1, l, r); else if(l > mid) return ask_mn(p<<1|1, l, r); else return min(ask_mn(p<<1, l, mid), ask_mn(p<<1|1, mid+1, r)); } inline ll fd(ll x, int L, int R) { if(L == R) return L; int mid =(L + R) >> 1; if(ask_mn(1, L, mid) <= x) return fd(x, L, mid); else return fd(x, mid+1, R); } inline ll solve(ll x, int L, int R){ while(ask_mn(1, L, R) <= x) x %= a[fd(x, L, R)]; return x; } int main() { n = readint(), q = readint(); for(int i = 1; i <= n; ++i) a[i] = readll(); build(1, 1, n); while(q--) { ll x; int L, R; x = readll(), L = readint(), R = readint(); write(solve(x, L, R)); puts(""); } return 0; }
解法3:
#include <cstdio> #include <algorithm> #include <iostream> #define pb(x) push_back(x) #define rg register const int maxn = 2e5 + 100; typedef unsigned long long ll; inline int readint(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } inline ll readll(){ char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } inline void write(ll x){ if(x>=10LL) write(x/10LL); putchar('0'+x%10LL); } using namespace std; int n, q; ll a[maxn], st[maxn][25]; inline void RMQ_init(){ for(int i = 0; i <= n; ++i) st[i][0] = a[i]; for(int j = 1; (1<<j) <= n; ++j){ for(int i = 1; i+(1<<j)-1 <= n; ++i){ st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]); } } } inline ll RMQ(int u, int v){ int k = (int)(log(v-u+1.0)/log(2.0)); return min(st[u][k], st[v-(1<<k)+1][k]); } inline ll fd(ll x, int L, int R) { if(L == R) return L; int mid =(L + R) >> 1; if(RMQ(L, mid) <= x) return fd(x, L, mid); else return fd(x, mid+1, R); } inline ll solve(ll x, int L, int R){ while(RMQ(L, R) <= x) x %= a[fd(x, L, R)]; return x; } int main() { n = readint(), q = readint(); for(int i = 1; i <= n; ++i) a[i] = readll(); RMQ_init(); while(q--) { ll x; int L, R; x = readll(), L = readint(), R = readint(); write(solve(x, L, R)); puts(""); } return 0; }
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