求用几句话把《题破山寺后禅院》和《黄鹤楼》串起来!是题破山寺后禅院和登岳阳楼ಥ_ಥ_百度作业帮
求用几句话把《题破山寺后禅院》和《黄鹤楼》串起来!是题破山寺后禅院和登岳阳楼ಥ_ಥ
求用几句话把《题破山寺后禅院》和《黄鹤楼》串起来!是题破山寺后禅院和登岳阳楼ಥ_ಥ
题破山寺后禅院常建清晨入古寺,初日照高林.曲径通幽处,禅房花木深.山光悦鸟性,潭影空人心.万籁此都寂,但余钟磬音.黄鹤楼崔颢昔人已乘黄鹤去,此地空余黄鹤楼.黄鹤一去不复返,白云千载空悠悠.晴川历历汉阳树,芳草萋萋鹦鹉洲.日暮乡关何处是?烟波江上使人愁.我们刚才听完了《题破山寺后禅院》这首诗的朗诵,这首诗使大家感受到了人间美的存在,但是如此美景,却无人共赏,自然是有些遗憾了.可惜好花常在,知音难共,崔颢的《黄鹤楼》就表达了对朋友的不舍.下面请大家一起欣赏一下《黄鹤楼》这首诗.
果咩 是题破山寺后禅院和登岳阳楼 能再帮忙想一下吗？会加分的
我们刚才听完了《题破山寺后禅院》这首诗的朗诵，这首诗使大家感受到了山间美的存在，不过天下间的美并不只有山间才存在，一些水边也存在着美景，洞庭湖畔的岳阳楼就是其中之一，下面请大家一起欣赏一下杜甫的《登岳阳楼》。MAX3237E | RS-232 Transceivers | Interface | Description & parametrics
|C0|0|0|0|2|0|3
&&&&&&&&&&&&
3-V To 5.5-V Single Channel RS-232 1 Mbit/s Line Driver/Receiver
In English
日本語表示
Tools & software
Description
Parametrics
Related end equipment
Companion products
Recommended alternative parts
&-&The device is an EXACT EQUIVALENT in functionality and parametrics to the compared device.&
Description
The MAX3237E consists of five line drivers, three line receivers, and a dual charge-pump circuit with &15-kV ESD protection pin to pin (serial-port connection pins, including GND). The device meets the requirements of TIA/EIA-232-F and provides the electrical interface between an asynchronous communication controller and the serial-port connector. The charge pump and four small external capacitors allow operation from a single 3-V to 5.5-V supply. This device operates at data signaling rates of 250 kbit/s in normal operating mode (MBAUD&=&GND) and 1Mbit/s when MBAUD = VCC. The driver output slew rate is a maximum of 30 V/&s.
The MAX3237E transmitters are disabled and the outputs are forced into high-impedance state when the device is in shutdown mode (SHDN = GND) and the supply current falls to less than 1 &A. Also, during shutdown, the onboard cha V+ is lowered to VCC, and V& is raised toward GND. Receiver outputs also can be placed in the high-impedance state by setting enable (EN) high. ROUT1B remains active all the time, regardless of the EN and SHDN condition.
The MAX3237EC is characterized for operation from 0&C to 70&C. The MAX3237EI is characterized for operation from &40&C to 85&C.
Meets or Exceeds the Requirements of TIA/EIA-232-F and ITU v.28 Standards Operates With 3-V to 5.5-V VCC Supply Operates From 250 kbits/s to 1 Mbit/s Low Standby Current . . . 1 &A Typical External Capacitors . . . 4 & 0.1 &F Accepts 5-V Logic Input With 3.3-V Supply Designed to Be Interchangeable With Maxim MAX3237E Latch-Up Performance Exceeds 100 mA Per JESD 78, Class II ESD Protection for RS-232 I/O Pins &15 kV & Human-Body Model (HBM) &8 kV & IEC, Contact Discharge &15 kV & IEC, Air-Gap Discharge APPLICATIONS Battery-Powered, Hand-Held, and Portable Equipment PDAs and Palmtop PCs Notebooks, Sub-Notebooks, and Laptops Digital Cameras Mobile Phones and Wireless Devices
Parametrics
Drivers Per Package
Receivers Per Package
Operating Temperature Range
Pin/Package
Approx. Price (US$)
Power Saving Feature
IEC- SUPPORT
1.00 | 1ku
0.88 | 1ku
Auto-Powerdown PlusDescription
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N - 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v
Change the weight of the ith edge to v
NEGATE a b
Negate the weight of every edge on the path from a to b
Find the maximum weight of edges on the path from a to b
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N - 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.
For each “QUERY” instruction, output the result on a separate line.
CHANGE 1 3
1&31 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONE
Sample Output
这道昨天坑了一天的题今天又坑了我2h，昨天是因为想不明白，还情有可原，今天竟然是因为数组没清！！！！！！！！！！！照着别人的程序一个一个函数换，换到最后发现竟然是建树函数错了……
好了，总结一下吧。
这次点里需要存的是边权，一条边的边权赋给深度大的点，因为1点无入边，所以进行线段树操作的时候要排除，线段树要维护最大值和最小值，执行negate操作时把max和min交换，需要延迟标记，总之还是姿势比较正常的线段树。
最后，要是用build函数，记得每次清空保存边权的数组！！！！！！！！！！！！
#include&iostream&
#include&cstring&
#include&cstdio&
#include&algorithm&
#define maxn 10010
#define lson l,m,rt&&1
#define rson m+1,r,rt&&1|1
int t,n,x,y,z,
int a[maxn][3],num[maxn];
char s[10];
struct edge{
}e[maxn&&1];
int h[maxn],
int siz[maxn],dep[maxn],fa[maxn],son[maxn],tid[maxn],rank[maxn],top[maxn];
int ma[maxn&&2],mi[maxn&&2];
bool flag[maxn&&2];
void ae(int u,int v)
e[++tp].to=v;
e[tp].next=h[u];
void Init()
memset(son,-1,sizeof(son));
memset(h,0,sizeof(h));
memset(ma,0,sizeof(ma));
memset(mi,0,sizeof(mi));
memset(flag,0,sizeof(flag));
memset(num,0,sizeof(num));
tp=0;cnt=0;
void dfs1(int u,int f,int d)
for(int i=h[u];e[i].i=e[i].next)
int v=e[i].
dfs1(v,u,d+1);
siz[u]+=siz[v];
if(son[u]==-1||siz[son[u]]&siz[v]) son[u]=v;
void dfs2(int u,int tp)
rank[tid[u]]=u;
if(son[u]==-1)
dfs2(son[u],tp);
for(int i=h[u];e[i].i=e[i].next)
int v=e[i].
if(v!=fa[u]&&v!=son[u]) dfs2(v,v);
void changev(int rt)
int temp=ma[rt];
ma[rt]=(-1)*mi[rt];
mi[rt]=(-1)*
void pushup(int rt)
ma[rt]=max(ma[rt&&1],ma[rt&&1|1]);
mi[rt]=min(mi[rt&&1],mi[rt&&1|1]);
void pushdown(int rt)
if(flag[rt])
flag[rt&&1]^=1;changev(rt&&1);
flag[rt&&1|1]^=1;changev(rt&&1|1);
flag[rt]=0;
void build(int l,int r,int rt)
ma[rt]=mi[rt]=num[rank[l]];
int m=(l+r)&&1;
build(lson);
build(rson);
pushup(rt);
void update(int x,int v,int l,int r,int rt)
ma[rt]=mi[rt]=v;
flag[rt]=0;
pushdown(rt);
int m=(l+r)&&1;
if(x&=m) update(x,v,lson);
else update(x,v,rson);
pushup(rt);
void ngt(int L,int R,int l,int r,int rt)
if(L&=l&&R&=r)
flag[rt]^=1;
changev(rt);
pushdown(rt);
int m=(l+r)&&1;
if(L&=m) ngt(L,R,lson);
if(R&m) ngt(L,R,rson);
pushup(rt);
int query(int L,int R,int l,int r,int rt)
if(L&=l&&R&=r)
return ma[rt];
pushdown(rt);
int ret=-0x3f3f3f;
int m=(l+r)&&1;
if(L&=m) ret=max(ret,query(L,R,lson));
if(R&m) ret=max(ret,query(L,R,rson));
pushup(rt);
void change(int a,int b)
while(top[a]!=top[b])
if(dep[top[a]]&dep[top[b]]) swap(a,b);
ngt(tid[top[a]],tid[a],2,n,1);
a=fa[top[a]];
if(dep[a]&dep[b]) swap(a,b);
ngt(tid[son[a]],tid[b],2,n,1);
int get_max(int a,int b)
if(a==b) return 0;
int ret=-0x3f3f3f3f;
while(top[a]!=top[b])
if(dep[top[a]]&dep[top[b]]) swap(a,b);
ret=max(ret,query(tid[top[a]],tid[a],2,n,1));
a=fa[top[a]];
if(dep[a]&dep[b]) swap(a,b);
ret=max(ret,query(tid[son[a]],tid[b],2,n,1));
int main()
//freopen("3.out","w",stdout);
while(t--)
scanf("%d",&n);
for(int i=1;i&n;i++)
scanf("%d%d%d",&x,&y,&z);
a[i][0]=x;
a[i][1]=y;
a[i][2]=z;
dfs1(1,0,0);
dfs2(1,1);
for(int i=1;i&n;i++)
if(dep[a[i][0]]&dep[a[i][1]]) num[a[i][0]]=a[i][2];
else num[a[i][1]]=a[i][2],a[i][0]=a[i][1];
build(2,n,1);
while(scanf("%s",s)&&s[0]!='D')
scanf("%d%d",&x,&y);
if(s[0]=='C') update(tid[a[x][0]],y,2,n,1);
else if(s[0]=='N') change(x,y);
else printf("%d\n",get_max(x,y));
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201
#include&iostream&#include&cstring&#include&cstdio&#include&algorithm&#define maxn 10010#define lson l,m,rt&&1#define rson m+1,r,rt&&1|1using namespace std;int t,n,x,y,z,cnt;int a[maxn][3],num[maxn];char s[10];struct edge{ int to; int next;}e[maxn&&1];int h[maxn],tp;int siz[maxn],dep[maxn],fa[maxn],son[maxn],tid[maxn],rank[maxn],top[maxn];int ma[maxn&&2],mi[maxn&&2];bool flag[maxn&&2];void ae(int u,int v){ e[++tp].to=v; e[tp].next=h[u]; h[u]=tp;}void Init(){ memset(son,-1,sizeof(son)); memset(h,0,sizeof(h)); memset(ma,0,sizeof(ma)); memset(mi,0,sizeof(mi)); memset(flag,0,sizeof(flag)); memset(num,0,sizeof(num)); tp=0;cnt=0;}void dfs1(int u,int f,int d){ siz[u]=1; dep[u]=d; fa[u]=f; for(int i=h[u];e[i].to;i=e[i].next) {
int v=e[i].to;
dfs1(v,u,d+1);
siz[u]+=siz[v];
if(son[u]==-1||siz[son[u]]&siz[v]) son[u]=v;
} } return;}void dfs2(int u,int tp){ top[u]=tp; tid[u]=++cnt; rank[tid[u]]=u; if(son[u]==-1) return; dfs2(son[u],tp); for(int i=h[u];e[i].to;i=e[i].next) {
int v=e[i].to;
if(v!=fa[u]&&v!=son[u]) dfs2(v,v); } return;}void changev(int rt){ int temp=ma[rt]; ma[rt]=(-1)*mi[rt]; mi[rt]=(-1)*temp;}void pushup(int rt){ ma[rt]=max(ma[rt&&1],ma[rt&&1|1]); mi[rt]=min(mi[rt&&1],mi[rt&&1|1]);}void pushdown(int rt){ if(flag[rt]) {
flag[rt&&1]^=1;changev(rt&&1);
flag[rt&&1|1]^=1;changev(rt&&1|1);
flag[rt]=0; }}void build(int l,int r,int rt){ if(l==r) {
ma[rt]=mi[rt]=num[rank[l]];
return; } int m=(l+r)&&1; build(lson); build(rson); pushup(rt);}void update(int x,int v,int l,int r,int rt){ if(l==r) {
ma[rt]=mi[rt]=v;
flag[rt]=0;
return; } pushdown(rt); int m=(l+r)&&1; if(x&=m) update(x,v,lson); else update(x,v,rson); pushup(rt);}void ngt(int L,int R,int l,int r,int rt){ if(L&=l&&R&=r) {
flag[rt]^=1;
changev(rt);
return; } pushdown(rt); int m=(l+r)&&1; if(L&=m) ngt(L,R,lson); if(R&m) ngt(L,R,rson); pushup(rt);}int query(int L,int R,int l,int r,int rt){ if(L&=l&&R&=r) return ma[rt]; pushdown(rt); int ret=-0x3f3f3f; int m=(l+r)&&1; if(L&=m) ret=max(ret,query(L,R,lson)); if(R&m) ret=max(ret,query(L,R,rson)); pushup(rt); return ret;}void change(int a,int b){ if(a==b) return; while(top[a]!=top[b]) {
if(dep[top[a]]&dep[top[b]]) swap(a,b);
ngt(tid[top[a]],tid[a],2,n,1);
a=fa[top[a]]; } if(a==b) return; if(dep[a]&dep[b]) swap(a,b); ngt(tid[son[a]],tid[b],2,n,1);}int get_max(int a,int b){ if(a==b) return 0; int ret=-0x3f3f3f3f; while(top[a]!=top[b]) {
if(dep[top[a]]&dep[top[b]]) swap(a,b);
ret=max(ret,query(tid[top[a]],tid[a],2,n,1));
a=fa[top[a]]; } if(a==b) return ret; if(dep[a]&dep[b]) swap(a,b); ret=max(ret,query(tid[son[a]],tid[b],2,n,1)); return ret;}int main(){ //freopen("3.out","w",stdout); cin&&t; while(t--) {
scanf("%d",&n);
for(int i=1;i&n;i++)
scanf("%d%d%d",&x,&y,&z);
a[i][0]=x;
a[i][1]=y;
a[i][2]=z;
dfs1(1,0,0);
dfs2(1,1);
for(int i=1;i&n;i++)
if(dep[a[i][0]]&dep[a[i][1]]) num[a[i][0]]=a[i][2];
else num[a[i][1]]=a[i][2],a[i][0]=a[i][1];
build(2,n,1);
while(scanf("%s",s)&&s[0]!='D')
scanf("%d%d",&x,&y);
if(s[0]=='C') update(tid[a[x][0]],y,2,n,1);
else if(s[0]=='N') change(x,y);
else printf("%d\n",get_max(x,y));
} } return 0;}GB/T 标准下载 GB/T
信息处理交换用穿孔纸带的盘心尺寸-标准下载网-www.bzxz.net
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信息处理交换用穿孔纸带的盘心尺寸
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信息处理交换用穿孔纸带的盘心尺寸
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信息处理交换用穿孔纸带的盘心尺寸
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标准简介：
本标准规定了卷绕穿孔纸带用的盘心尺寸。本标准规定的盘心尺寸只适用于宽度为25.4mm的穿孔纸带。 GB/T
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