【DP】 TOJ 4103. Chinese Hockey 3

注意到只要是递减数列，总和为总分数，都符合题目意义。。。dp暴力打表，交表即可。。。

#include <bits/stdc++.h>#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 100005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headLL dp[4000][200][55];int n;LL dfs(int sum, int pre, int pos){if(dp[sum][pre][pos] != -1) return dp[sum][pre][pos];if(pos == n + 1) {if(sum == 3 * n * (n - 1) / 2) return true;else return false;}int limit = 3 * pos * (n - pos) + pos * (pos - 1) * 3 / 2 - sum;LL ans = 0;for(int i = 0; i <= pre && i <= limit; i++) {ans = (ans + dfs(sum + i, i, pos + 1)) % mod;}return dp[sum][pre][pos] = ans % mod;}int work(){memset(dp, -1, sizeof dp);return dfs(0, (n - 1) * 3, 1);}int main(){//freopen("output", "w", stdout);for(int i = 1; i <= 50; i++) {n = i;printf("if(n == %d) ", i);printf("printf(\"%lld\");", work());printf("\n");}return 0;}

#include <stdio.h>using namespace std;void work(){int n;scanf("%d", &n);if(n == 1) printf("1");if(n == 2) printf("2");if(n == 3) printf("8");if(n == 4) printf("37");if(n == 5) printf("198");if(n == 6) printf("1178");if(n == 7) printf("7548");if(n == 8) printf("50944");if(n == 9) printf("357855");if(n == 10) printf("2595250");if(n == 11) printf("19313372");if(n == 12) printf("146815503");if(n == 13) printf("136158488");if(n == 14) printf("927025933");if(n == 15) printf("65653738");if(n == 16) printf("215408350");if(n == 17) printf("746589264");if(n == 18) printf("724066327");if(n == 19) printf("631274816");if(n == 20) printf("422001336");if(n == 21) printf("7146650");if(n == 22) printf("407217283");if(n == 23) printf("393518927");if(n == 24) printf("965151303");if(n == 25) printf("895251280");if(n == 26) printf("864749285");if(n == 27) printf("934373752");if(n == 28) printf("467108499");if(n == 29) printf("418735743");if(n == 30) printf("451847922");if(n == 31) printf("890597718");if(n == 32) printf("824730990");if(n == 33) printf("739940095");if(n == 34) printf("858720713");if(n == 35) printf("600431442");if(n == 36) printf("577631035");if(n == 37) printf("867745635");if(n == 38) printf("3561686");if(n == 39) printf("302514002");if(n == 40) printf("527947240");if(n == 41) printf("805543258");if(n == 42) printf("150530321");if(n == 43) printf("534209877");if(n == 44) printf("85487795");if(n == 45) printf("939816214");if(n == 46) printf("315353659");if(n == 47) printf("496074452");if(n == 48) printf("538654703");if(n == 49) printf("326456510");if(n == 50) printf("82203529");printf("\n");}int main(){int _;scanf("%d", &_);while(_--) work();}