BZOJ2118: 墨墨的等式(最短路 数论)
作者:互联网
题意
墨墨突然对等式很感兴趣,他正在研究a1x1+a2y2+…+anxn=B存在非负整数解的条件,他要求你编写一个程序,给定N、{an}、以及B的取值范围,求出有多少B可以使等式存在非负整数解。
Sol
maya神仙题啊,感觉自己做题难度跨度太大了qwq。
这里有一篇讲的非常好的博客https://blog.csdn.net/w4149/article/details/66476606?locationNum=3&fps=1
思路大概就是 利用取余的性质,把能够构造出来的解表示成统一的形式
发现该形式可以通过最短路更新
然后就做完了。。
#include<cstdio> #include<algorithm> #include<stack> #include<queue> #include<cmath> #include<cstring> #define lb(x) (x & (-x)) #define Pair pair<int, int> #define fi first #define se second #define MP(x, y) make_pair(x, y) #define LL long long using namespace std; const int MAXN = 1e6 + 10; inline LL read() { 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; } int N, vis[MAXN]; LL dis[MAXN], a[MAXN], Mi = 1e15, Bmin, Bmax; void SPFA() { queue<int> q; memset(dis, 0xf, sizeof(dis)); dis[0] = 0; vis[0] = 1, q.push(0); while(!q.empty()) { int p = q.front(); q.pop(); vis[p] = 1; for(int i = 1; i <= N; i++) { int to = (p + a[i]) % Mi;//tag if(dis[to] > dis[p] + a[i]) { dis[to] = dis[p] + a[i]; if(!vis[to]) vis[to] = 1, q.push(to); } } } } LL Query(LL x) { LL rt = 0; for(int i = 0; i < Mi; i++) if(dis[i] <= x) rt += (x - dis[i]) / Mi + 1; return rt; } int main() { N = read(); Bmin = read(); Bmax = read(); for(int i = 1; i <= N; i++) { a[i] = read(); if(a[i] != 0) Mi = min(Mi, a[i]); } SPFA(); printf("%lld", Query(Bmax) - Query(Bmin - 1)); return 0; } /* */
标签:BZOJ2118,int,短路,dis,vis,墨墨,include,LL,define 来源: https://blog.51cto.com/u_15239936/2868697