题目链接
题解
orz竟然是最短路
我们去\(0\)后取出最小的\(a[i]\)，记为\(p\)，然后考虑模\(p\)下的\(B\)
一个数\(i\)能被凑出，那么\(i + p\)也能被凑出
所以我们只需找出最小的凑出\(i\)的代价
我们如果将同余下的和看作点，那么加上一个数就相当于在点间转移的边
所以我们只需跑最短路即可求出每个\(i\)的最小代价，然后就可以计算\(Bmin\)和\(Bmax\)以内分别有多少个\(i\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 5000005;
const LL INF = 100000000000000001ll;
inline LL read(){
LL out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
struct node{
int u; LL d;
};
inline bool operator <(const node& a,const node& b){
return a.d > b.d;
}
inline bool operator ==(const node& a,const node& b){
return a.u == b.u && a.d == b.d;
}
struct Heap{
priority_queue<node> a,b;
void ck(){while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();}
int size(){return a.size() - b.size();}
node top(){ck(); node x = a.top(); a.pop(); return x;}
void del(node x){ck(); b.push(x);}
void ins(node x){ck(); a.push(x);}
}H;
int N,a[maxn],P;
LL d[maxn]; int vis[maxn];
int h[maxn],ne;
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
}
void work(){
for (int i = 0; i < P; i++){
for (int j = 1; j <= N; j++)
build(i,(i + a[j]) % P,a[j]);
}
for (int i = 1; i < P; i++) d[i] = INF;
d[0] = 0; H.ins((node){0,d[0]}); vis[0] = true;
node u;
while (H.size()){
u = H.top();
Redge(u.u) if (!vis[to = ed[k].to] && d[to] > d[u.u] + ed[k].w){
if (d[to] != INF) H.del((node){to,d[to]});
d[to] = d[u.u] + ed[k].w;
H.ins((node){to,d[to]});
}
}
}
int main(){
N = read(); LL L = read(),R = read(); P = INF;
REP(i,N){
a[i] = read();
if (!a[i]) i--,N--;
}
if (!N){
if (L) puts("0");
else puts("1");
return 0;
}
REP(i,N) P = min(P,a[i]);
work();
L--;
LL ansl = 0,ansr = 0;
for (int i = 0; i < P; i++){
if (d[i] <= L){
ansl++;
ansl += (L - d[i]) / P;
}
if (d[i] <= R){
ansr++;
ansr += (R - d[i]) / P;
}
}
printf("%lld\n",ansr - ansl);
return 0;
}