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  • 洛谷P4588 [TJOI2018]数学计算 【线段树】

    题目链接

    洛谷P4588

    题解

    用线段树维护即可

    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #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
    #define ls (u << 1)
    #define rs (u << 1 | 1)
    using namespace std;
    const int maxn = 100005,maxm = 100005,INF = 1000000000;
    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;
    }
    int Q,P;
    LL val[maxn << 2];
    void build(int u,int l,int r){
    	if (l == r){val[u] = 1; return;}
    	int mid = l + r >> 1;
    	build(ls,l,mid);
    	build(rs,mid + 1,r);
    	val[u] = 1ll * val[ls] * val[rs] % P;
    }
    void modify(int u,int l,int r,int pos,LL v){
    	if (l == r){val[u] = v; return;}
    	int mid = l + r >> 1;
    	if (mid >= pos) modify(ls,l,mid,pos,v);
    	else modify(rs,mid + 1,r,pos,v);
    	val[u] = 1ll * val[ls] * val[rs] % P;
    }
    LL query(int u,int l,int r,int L,int R){
    	if (l >= L && r <= R) return val[u];
    	int mid = l + r >> 1;
    	if (mid >= R) return query(ls,l,mid,L,R);
    	if (mid < L) return query(rs,mid + 1,r,L,R);
    	return 1ll * query(ls,l,mid,L,R) * query(rs,mid + 1,r,L,R) % P;
    }
    int main(){
    	int T = read();
    	while (T--){
    		Q = read(); P = read(); LL opt,v;
    		build(1,1,Q);
    		REP(i,Q){
    			opt = read(); v = read();
    			if (opt & 1) modify(1,1,Q,i,v % P);
    			else modify(1,1,Q,v,1);
    			printf("%lld
    ",query(1,1,Q,1,i));
    		}
    	}
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9050414.html
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