UVA10616 - Divisible Group Sums(dp)
2014-11-16 21:47
267 查看
UVA10616 - Divisible Group Sums(dp)
题目链接
题目大意:N个数,选择m个数出来,问相加的和能够整除MOD有多少种选择方式。
解题思路:从1到N数选择过去,每个数有选和不选两种可能,并且(num + d) % MOD = num % MOD + d % MOD, 所以可以这么做,最后判断一下余数等于0么。坑点是这题N个数会有负数,负数的取模 (num % MOD + MOD) % MOD.这一题因为没有控制选择的数目m的增加次数,导致re了好久,简直无语死了。。。
代码:
题目链接
题目大意:N个数,选择m个数出来,问相加的和能够整除MOD有多少种选择方式。
解题思路:从1到N数选择过去,每个数有选和不选两种可能,并且(num + d) % MOD = num % MOD + d % MOD, 所以可以这么做,最后判断一下余数等于0么。坑点是这题N个数会有负数,负数的取模 (num % MOD + MOD) % MOD.这一题因为没有控制选择的数目m的增加次数,导致re了好久,简直无语死了。。。
代码:
#include <cstdio> #include <cstring> typedef long long ll; const int maxn = 205; const int maxm = 15; const int maxd = 25; int N, Q, MOD, M; int num[maxn], tmp[maxn]; ll f[maxn][maxm][maxd]; void init () { for (int i = 0; i < N; i++) scanf ("%d", &num[i]); } ll dp (int n, int m, int d) { ll& ans = f [m][d]; if (ans != -1) return ans; if (n == N) { if (m == M && d == 0) return ans = 1; return ans = 0; } ans = 0; if (m < M) ans += dp(n + 1, m + 1, (d + tmp ) % MOD); ans += dp(n + 1, m, d); return ans; } int main () { int cas = 0; while (scanf ("%d%d", &N, &Q) && (N || Q)) { init(); printf ("SET %d:\n", ++cas); for (int i = 0; i < Q; i++) { scanf ("%d%d", &MOD, &M); memset (f, -1, sizeof (f)); for (int j = 0; j < N; j++) tmp[j] = (num[j] % MOD + MOD) % MOD; printf ("QUERY %d: %lld\n", i + 1, dp(0, 0, 0)); } } return 0; }
相关文章推荐
- UVa 10616 Divisible Group Sums (DFS&DP)
- UVa - 10616 - Divisible Group Sums(dp)
- 10616 - Divisible Group Sums(dp背包)
- UVa 10616 - Divisible Group Sums
- UVa 10616 - Divisible Group Sums
- uva 10616 - Divisible Group Sums(计数)
- UVA 10616 Divisible Group Sums
- Light oj 1125 - Divisible Group Sums(简单dp)
- LightOJ1125 Divisible Group Sums(DP)
- LightOJ1125 - Divisible Group Sums - 背包dp
- LightOJ 1125 Divisible Group Sums (dp)
- Light OJ 1125 Divisible Group Sums 背包DP 2017/1/23
- LightOJ - 1125 Divisible Group Sums(DP)
- Light OJ 1125 Divisible Group Sums (DP)
- Light oj 1125 - Divisible Group Sums (dp)
- 1125 - Divisible Group Sums (DP)
- lightoj 1125 - Divisible Group Sums
- LightOJ1125 Divisible Group Sums
- Light OJ 1125 Divisible Group Sums
- Divisible Group Sums LightOJ - 1125 变形01背包