2015 Nanyang Ccpc c-the Battle of Chibi DP

C-the Battle of Chibi

Time Limit:1 Sec

Memory limit:256 MB

Topic Connection

No

Description

Cao Cao made up a big army and is going to invade the whole South China. Yu Zhou was worried about it. He thought the only-to-beat Cao Cao is to a spy in Cao Cao ' s army. But all generals and soldiers of the Cao Cao were loyal, it ' s impossible to convince any of the them to betray Cao Cao.

So there are only one-of-a-Zhou, send someone to fake surrender Cao Cao. Gai Huang is selected for this important mission. However, Cao Cao is not easy-to-believe others, so Gai Huang must leak some important information to Cao Cao before Surre Ndering.

Yu Zhou discussed with Gai Huang and worked off N information to being leaked, in happening order. Each of the information is estimated to have AI value in Cao Cao ' s opinion.

Actually, if you leak information with strict increasing value could accelerate making Cao Cao believe you. So Gai Huang decided to leak exact M information with strict increasing value in happening order. In other words, Gai Huang won't change the order of the N information and just select M of them. Find out how many ways Gai Huang could does this.

Inputthe first line of the input gives the number of test cases, T (1≤100). T test Cases follow.

Each test case begins with numbers N (1≤n≤103) and M (1≤m≤n), indicating the number of information and number of Informa tion Gai Huang would select. Then N numbers on a line, the ith number AI (1≤ai≤109) indicates the value in Cao Cao's opinion of the ith information in H Appening order. Outputfor each test case, output one line containing case #x: Y, where x is the test Case number (starting from 1) and Y-is The ways Gai Huang can select the information.

The result is too large, and you need to the output of the result mod by 1000000007 (109+7).Sample Input

23 21 2 33 23 2 1

Sample Output

Case #1:3
Case #2:0

HINT

Test instructions

Give you the number of n, and find out how many ascending sub-sequences with a length of M

Exercises

N^3 would have been a direct force, but it would be tle.

So it's better to use a tree-like array to N^2logn.

Code:

#include <iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<queue>#include<map>#include<Set>#include<vector>#include<string>#include<stack>#include<bitset>#defineINF 1000000005#defineEPS 1e-12#definePI ACOs (-1.0)#defineLL Long Longusing namespacestd;Const intMAXN =1005;Const intMoD =1000000007; inlineLong LongRead () {Long Longx=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}intA[MAXN], B[MAXN], C[MAXN][MAXN], N, m, Dp[maxn][maxn];inlinevoidInit () { for(inti =1; I <= N; i++) {A[i]= B[i] =0;  for(intj =1; J <= N; J + +) Dp[i][j]= C[i][j] =0; }    return;} InlineintLowbit (intx) {    returnX & (-x);} InlinevoidUpdate (intLintPosintkey) {     while(Pos <=N) {C[l][pos]= (C[l][pos] +key);  while(c[l][pos]>=MoD) C[l][pos]-=MoD; POS= pos +Lowbit (POS); }    return;} InlineintSum (intLintPOS) {    inttemp =0;  while(POS) {temp= (temp +C[l][pos]);  while(temp >= MoD) Temp-=MoD; POS= pos-Lowbit (POS); }    returntemp;}intMain () {intT, CAS =0; scanf ("%d", &u);  while(t--) {scanf ("%d%d", &n, &m);        Init ();  for(inti =1; I <= N; i++) {A[i]=read (); //scanf ("%d", &a[i]);B[i] =A[i]; } sort (b+1, B +1+N);  for(inti =1; I <= N; i++) A[i]= Lower_bound (b +1, B +1+ N, a[i])-b; //cout <<-1 << endl;         for(inti =1; I <= N; i++)        {           //cout << a[i] << Endl;dp[i][1] =1;  for(intj =2; J <= I; J + +) Dp[i][j]= Sum (J-1, A[i]-1);  for(intj =1; J <= I; J + +) Update (J, A[i], dp[i][j]); }        Long LongAns =0;  for(inti =1; I <= N; i++) {ans= (ans +dp[i][m]);  while(ans>=MoD) ans-=mod;//% MoD; //cout << dp[i][m] << Endl;} printf ("Case #%d:%lld\n", ++cas, ans); }    return 0;}

2015 Nanyang Ccpc c-the Battle of Chibi DP