hdu 4902——Nice boat

来源：互联网发布：程序员常用的工具编辑：程序博客网时间：2024/06/07 03:30

题意：维护一个线段树，有两种更新，一种是修改一个区间的值。一种是在一个区间内把所有大于x的数改成这个数与x的最大公约数。
代码如下

#include<iostream>#include<cstring>#include<vector>#include<algorithm>#include<cstdio>#include<map>#include<cmath>#include<assert.h>#include<iomanip>#include<assert.h>using namespace std;typedef long long ll;const int maxn = 400005;int gcd(int a,int b){    int c = a%b;    for(;c!=0;c=a%b){        a = b;        b = c;    }    return b;}int valv[maxn];int maxv[maxn];void pushdown(int l,int r,int o){    if(l == r)return;    if(valv[o]==-1)return;    valv[o*2] = valv[o];    valv[o*2+1] = valv[o];    maxv[o*2] = valv[o*2];    maxv[o*2+1] = valv[o*2+1];    valv[o] = -1;}int yl,yr;int v;void update1(int l,int r,int o){    if(yl<=l&&yr>=r){        valv[o] = v;        maxv[o] = valv[o];        return;    }    int m = (l+r);m>>=1;    pushdown(l,r,o);    if(yl<=m){        update1(l,m,o*2);    }    if(yr>m){        update1(m+1,r,o*2+1);    }    maxv[o] = max(maxv[o*2],maxv[o*2+1]);}void update2(int l,int r,int o){    if(maxv[o]<v)return;    if(yl<=l&&yr>=r&&valv[o]>=0){        valv[o] = gcd(valv[o],v);        maxv[o] = valv[o];        return;    }    pushdown(l,r,o);    int m = l+r;m>>=1;    if(yl<=m){        update2(l,m,o*2);    }    if(yr>m){        update2(m+1,r,o*2+1);    }    maxv[o] = max(maxv[o*2],maxv[o*2+1]);}void solve(int l,int r,int o){    int m = l+r;m>>=1;    if(valv[o] >=0){        for(int i=l;i<=r;++i){            printf("%d ",valv[o]);        }        return;    }    solve(l,m,o*2);    solve(m+1,r,o*2+1);}int main(){//    freopen("data.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--){        memset(maxv,-1,sizeof(maxv));        memset(valv,-1,sizeof(valv));        int n;        scanf("%d",&n);        for(int i=1;i<=n;++i){            int a;            scanf("%d",&a);            yl = i;yr = i;            v = a;            update1(1,n,1);        }        int q;        scanf("%d",&q);        while(q--){            int t;            scanf("%d%d%d%d",&t,&yl,&yr,&v);            if(t == 1)                update1(1,n,1);            else {                update2(1,n,1);            }        }        solve(1,n,1);        puts("");    }    return 0;}

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