2559. [NOIP2016] combination number problem, 2559noip2016

2559. [NOIP2016] combination number problem, 2559noip2016
[Description]

 

[Input format]

Read data from a file.

The first line has two integers, t and k. t represents the total number of test data sets in the test point. For the meaning of k, see [Problem description ].

Next, there are two integers n and m in each row in Line t. For the meanings of n and m, see [Problem description ].

[Output format]

Output to the file.

T rows. An integer in each row indicates the number of (I, j) pairs in all 0 <= I <= n, 0 <= j <= min (I, m) C (j, I) is a multiple of k.

[Example 1 input]

1 23 3

[Example 1 Output]

1

Tip]

In all possible cases, only C (1, 2) is a multiple of 2.

[Example 2 input]

2 54 56 7

[Example 2 output]

07

[Source]

NOIP2016 official data,

 

This question was awesome to me a few days ago.

But today we have learned a formula.

C, n + 1, m = C, n, m, n, m,

I just wanted to push it for a while. Later I pushed it to 55.

The value of % k is 90.

Later, I couldn't do it. I looked at the question of the great god. I could use the prefix and

So it seems that noip2016's question is not very difficult,

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 const int MAXN = 2001; 7 int T, k, n, m; 8 int dp [MAXN] [MAXN]; 9 int read (int & n) 10 {11 int flag = 0, x = 0; char c = '/'; 12 while (c <'0' | c> '9') {c = getchar (); if (c = '-') flag = 1;} 13 while (c> = '0' & c <= '9') x = x * 10 + (c-48), c = getchar (); 14 if (flag) n =-x; else n = x; 15} 16 int main () 17 {18 freopen ("problem. in "," r ", stdin); 19 freopen (" problem. out "," w ", stdout); 20 read (T); read (k); 21 for (int I = 0; I <= 2001; I ++) 22 dp [I] [0] = 1, dp [I] [I] = 1; 23 for (int I = 0; I <= 2001; I ++) 24 for (int j = 0; j <= 2001; j ++) 25 if (dp [I + 1] [j] = 0) 26 dp [I + 1] [j] = (dp [I] [j] + dp [I] [J-1]); 27 28 while (T --) 29 {30 read (n); read (m); 31 int ans = 0; 32 for (int I = 1; I <= n; I ++) 33 for (int j = 1; j <= min (I, m); j ++) 34 if (dp [I] [j] % k = 0) 35 ans ++; 36 printf ("% d \ n", ans); 37} 38 return 0; 39}55 1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # define LL long 6 using namespace std; 7 const int MAXN = 2003; 8 int T, k, n, m; 9 LL dp [MAXN] [MAXN]; 10 LL sum [MAXN] [MAXN]; 11 int read (int & n) 12 {13 int flag = 0, x = 0; char c = '/'; 14 while (c <'0' | c> '9') {c = getchar (); if (c = '-') flag = 1 ;} 15 while (c> = '0' & c <= '9') x = x * 10 + (c-48), c = getchar (); 16 if (flag) n =-x; else n = x; 17} 18 int main () 19 {20 freopen ("problem. in "," r ", stdin); 21 freopen (" problem. out "," w ", stdout); 22 23 read (T); read (k); 24 for (int I = 1; I <= 2002; I ++) 25 dp [I] [I] = 1, dp [I] [1] = I % k; 26 for (int I = 0; I <= 2002; I ++) 27 for (int j = 2; j <I; j ++) 28 dp [I] [j] = (dp [I-1] [j] % k + dp [I-1] [J-1] % k) % k; 29 30/* for (int l = 0; l <= 50; l ++) 31 for (int j = 0; j <= l; j ++) 32 cout <dp [l] [j]; 33 cout <endl; */34 // cout <dp [1] [2]; 35 36 for (int I = 1; I <= 2002; I ++) 37 for (int j = 1; j <= I; j ++) 38 {39 if (dp [I] [j] = 0) sum [I] [j] = sum [I] [J-1] + 1; // meet condition 40 else sum [I] [j] = sum [I] [J-1]; 41} 42 43 while (T --) 44 {45 read (n ); read (m); 46 LL ans = 0; 47 for (int I = 1; I <= n; I ++) 48 ans + = sum [I] [min (I, m)]; 49 cout <ans <endl; 50 // printf ("% d \ n ", sum [n] [min (n, m)]); 51} 52 return 0; 53}AC