Chunked Learning Notes

Recently, when I was doing a senior simulation, I found that there are a lot of parts that can be divided, so I learned to divide the pieces.

Sub-block is actually a way to do the idea, because I just learned a linear block, so light talk about the linear bar: In fact, the general idea is when we want to a larger size of the problem to modify or query, with data structure is not good to maintain, but the scope is too large and can not use violence to do, So we can divide this big data into about sqrt (n), but not necessarily.

Put together an example: a paper from the national team

Because it is the person who wrote the paper did not have the data, I took a shot, I think it should be a, if anyone has made a mistake, remember to tell me.

#include <iostream> #include <cstdio> #include <cstring>using namespace std; #include <cmath> int N,m,a[100010]={0},len,n=1,d[400]={0},num[400]={0},influence[400][400];char Ch;int Main () {freopen ("cin.in", "R" , stdin), Freopen ("A.out", "w", stdout), int i,j,x,y;scanf ("%d%d", &n,&m), for (I=1;i<=n;++i) scanf ("%d", &a[i]); len=sqrt (n), while (1) {n+=1;if (n*len==n) Break;else if (n*len>n) break;}  for (I=1;i<=n;++i) for (j=1;j<=len;++j) {if ((i-1) *len+j>n) break;  num[i]+=a[(i-1) *len+j]; }if (len==1) {while (m--) {int t,x,y;cin>>t>>x>>y;if (t==1) {for (j=1;j<=n;j+=x) a[j]+=y;} Else{int ans=0;for (j=x;j<=y;++j) Ans+=a[j];cout<<ans<<endl;}} Else{while (m--) {int t;scanf ("%d%d%d", &t,&x,&y), if (t==1) {if (X>=len) {i=1;while (i<=n) {if (i%len== 0) Num[i/len]+=y;else num[i/len+1]+=y;a[i]+=y;i+=x;}} else d[x]+=y;} Else{int ans=0,l,r,ll,rr;l= (x%len==0) X/len: (x/len+1); r= (y%len==0)? (Y/len):(y/len+1); ll=len* (L-1) +1==x?l:l+1;rr=y%len==0?R:r-1;for (i=ll;i<=rr;++i) ans+=num[i];if (LL&LT;=RR) {if (L&LT;LL) for (i=x;i<=l*len;++i) ans+=a[i];if (R&GT;RR) for (I=rr*len+1;i<=y;++i) ans+=a[i];} Else{for (I=x;i<=y;++i) ans+=a[i];}    for (i=1;i<=len;++i) {int r1,r2;    r1=x/i;    if (x%i>1) r1+=1;    r2=y/i;    if (y%i>=1) r2+=1;    ans+= (R2-R1) *d[i]; if (i==1) ans+=d[i];} printf ("%d\n", ans);   }}}}

Chunked Learning Notes