• **错误积累&&防止GG写法总结


    19.3.20 关于int与ll

    1.如果一道题时间充足,把全部变量定义成long long

    2.特别注意最上面宏定义的变量类型,特别是FOR

    3.如果是int,用1LL* ......

    **实例,让我崩溃

    1 #include<bits/stdc++.h>
    2 using namespace std;
    3 #define rg register
    4 typedef long long ll;
    5 #define gmax(a,b) a=max(a,b)
    6 #define gmin(a,b) a=min(a,b)
    View Code
    1 #define FOR(i,a,b) for(rg int i=a;i<=b;++i)
    2 #define For(i,a,b) for(rg int i=a;i>=b;--i)
     1 #define gc pa==pb&&(pb=(pa=buf)+fread(buf,1,100000,stdin),pa==pb)?EOF:*pa++
     2 static char buf[100000],*pa(buf),*pb(buf);
     3 inline int rd()
     4 {
     5     rg int x(0),w(1);
     6     rg char c(gc);
     7     while(c<'0' || c>'9')
     8     {
     9         if(c=='-') w=-1;
    10         c=gc;
    11     }
    12     while(c>='0' && c<='9') x=x*10+c-48,c=gc;
    13     return x*w;
    14 }
    15 
    16 const int N=1000005;
    17 const ll MOD=998244353,ny=499122177;
    18 ll f[N],g[N],tn[N],jc[N];
    19 int c[N],a[N],n;
    20 ll ans;
    21 
    22 void add(int x,int y)
    23 {
    24     for(;x<=n;x+=x&(-x)) c[x]+=y;
    25 }
    26 
    27 ll ask(int x)
    28 {
    29     ll res=0;
    30     for(;x;x-=x&(-x)) res+=c[x];
    31     return res;
    32 }
    33 
    34 int main()
    35 {
    36     freopen("sort.in","r",stdin);
    37     freopen("sort.out","w",stdout);
    38     n=rd();FOR(i,1,n) a[i]=rd();
    39     jc[0]=1;FOR(i,1,n) jc[i]=(i*jc[i-1])%MOD;
    View Code
    1 FOR(i,2,n) f[i]=((i*f[i-1])%MOD+(1LL*i*(i-1)/2)%MOD*jc[i-1]%MOD)%MOD;
    1     tn[1]=1;add(a[n],1);
    2     FOR(i,2,n)
    3         g[i]=ask(a[n-i+1]),add(a[n-i+1],1),
    4         tn[i]=((g[i]*jc[i-1])%MOD+tn[i-1])%MOD;
    5     FOR(i,1,n) ans=(ans+g[i]*f[i-1]%MOD+g[i]*(g[i]-1)%MOD*ny%MOD*jc[i-1]%MOD+g[i]*tn[i-1]%MOD)%MOD;
    6     cout<<ans;
    7     return 0;
    8 }
    View Code

    21.10.26

    队列是while(head < tail)

    不是小于等于!!!!!

    while ( hd < tl )
     1     {
     2         ++hd;
     3         xx=l[hd][0];yy=l[hd][1];
     4         FOR(i,0,7)
     5         {
     6             xp=xx+dx[i];yp=yy+dy[i];
     7             if((xp>=1)&&(xp<=n)&&(yp>=1)&&(yp<=m)&&(a[xp][yp]==-1))
     8             {
     9                 a[xp][yp]=a[xx][yy]+1;
    10                 ++tl;l[tl][0]=xp;l[tl][1]=yp;
    11             }
    12         } 
    13     }
    View Code

    21.11.15

    要注意防止while()大括号内可能出现不操作的情况造成死循环

    while(len>1) 
        if(x[len]==0) --len;
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  • 原文地址:https://www.cnblogs.com/universeplayer/p/10565654.html
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