1901:zju2112 Dynamic Rankings

1901:zju2112 Dynamic Rankings time limit:10 Sec Memory limit:128 MB
submit:7339 solved:3055
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Given a sequence of n A[1],a[2],a[3]......a[n], the program must answer this question: For a given i,j,k, the small number of K in A[i],a[i+1],a[i+2]......a[j] is (1≤k≤j-i+1), and, You can change the value of some a[i], and after the change, the program can continue to answer the above question for the changed a. You need to make a program that reads the sequence a from the input file and then reads in a series of instructions, including the Ask and modify instructions. For each inquiry instruction, you must output the correct answer. The first line has two positive integers n (1≤n≤10000), M (1≤m≤10000). Indicates the length of the sequence and the number of instructions, respectively. The second line has n number, which means a[1],a[2]......a[n], these numbers are smaller than 10^9. The next M-line describes each instruction, and the format of each line is one of the following two formats. Q I j K or C i t Q i j K (I,j,k is a number, 1≤i≤j≤n, 1≤k≤j-i+1) indicates an inquiry instruction asking for a small number of k in the A[i],a[i+1]......a[j]. C i T (1≤i≤n,0≤t≤10^9) means to change a[i] into T.

Input

For each query, you need to output his answer, with each output occupying a separate line.

Output Sample Input5 3
3 2 1) 4 7
Q 1 4 3
C 2 6
Q 2 5 3
Sample Output3
6
HINT

20% of the data, m,n≤100; 40% of the data, m,n≤1000; 100% of the data, m,n≤10000.

Source

#include <cstdio>#include<algorithm>using namespacestd;Const intn=1e5+5;Const intm=2200001;intn,m,cnt,top,a,b,x[n],a[n],b[n],v[n],num[n<<1],k[n],l[ -],r[ -];intRoot[n],ls[m],rs[m],sum[m];intLowbit (intx) {    returnx&-x;}voidInsertint&k,intLastintLintRintXintadd) {k=++CNT; intMid=l+r>>1; LS[K]=Ls[last]; RS[K]=Rs[last]; SUM[K]=sum[last]+add; if(L==R)return; if(x<=mid) Insert (LS[K],LS[LAST],L,MID,X,ADD); ElseInsert (rs[k],rs[last],mid+1, R,x,add);} intQueryintLintRintk) {    if(L==R)returnl; intMid=l+r>>1; intSuml=0, sumr=0;  for(intI=1; i<=a;i++) suml+=Sum[ls[l[i]];  for(intI=1; i<=b;i++) sumr+=Sum[ls[r[i]]; if(sumr-suml>=k) {         for(intI=1; i<=a;i++) l[i]=Ls[l[i]];  for(intI=1; i<=b;i++) r[i]=Ls[r[i]]; returnquery (L,MID,K); }    Else{         for(intI=1; i<=a;i++) l[i]=Rs[l[i]];  for(intI=1; i<=b;i++) r[i]=Rs[r[i]]; returnQuery (mid+1, r,k-(sumr-suml)); }}     intMain () {scanf ("%d%d", &n,&m); top=N;  for(intI=1; i<=n;i++) scanf ("%d", &v[i]), num[i]=V[i]; Charopt[6];  for(intI=1; i<=m;i++) {scanf ("%s%d%d",opt,&a[i],&B[i]); if(opt[0]=='Q') scanf ("%d", &x[i]), k[i]=1; Elsenum[++top]=B[i]; } sort (num+1, num+top+1); intTot=unique (num+1, num+top+1)-num-1;  for(intI=1; i<=n;i++){        intT=lower_bound (num+1, num+tot+1, V[i])-num;  for(intj=i;j<=n;j+=Lowbit (j)) Insert (Root[j],root[j],1, Tot,t,1); }     for(intI=1; i<=m;i++){        if(K[i]) {a=0, b=0; a[i]--;  for(intj=a[i];j>=1; J-=lowbit (j)) l[++a]=Root[j];  for(intj=b[i];j>=1; J-=lowbit (j)) r[++b]=Root[j]; printf ("%d\n", Num[query (1, Tot,x[i])); }        Else{            intT=lower_bound (num+1, num+tot+1, V[a[i]])-num;  for(intj=a[i];j<=n;j+=Lowbit (j)) Insert (Root[j],root[j],1, tot,t,-1); V[a[i]]=B[i]; T=lower_bound (num+1, num+tot+1, V[a[i]])-num;  for(intj=a[i];j<=n;j+=Lowbit (j)) Insert (Root[j],root[j],1, Tot,t,1); }    }}

1901:zju2112 Dynamic Rankings