Light oj 1125 - Divisible Group Sums (dp)

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1125

题意：

给你n个数，q次询问，每次询问问你取其中m个数是d的整数倍的方案数。

题意：

dp[i][j][k] 表示前i个数, %d=j, 取了k个的方案数。

ID	SUBMISSION TIME	PROBLEM	SOURCE	CPU	MEMORY	VERDICT
839200	2016-10-15 14:59:00	1125 - Divisible Group Sums	C++	0.012	1688	Accepted
839186	2016-10-15 14:35:08	1125 - Divisible Group Sums	C++	0.336	1688	Accepted

没优化前 0.336：

 #include <cstdio>
 #include <cstring>
 using namespace std;
 typedef long long LL;
 const int N = ;
 LL dp[][][]; //dp[i][j][k]表示包含i
 int a[N], b[N];

 void solve(int n, int m, int d) {
     memset(dp, , sizeof(dp));
     for(int i = ; i <= n; ++i) {
         dp[i][b[i]][] = ;
         for(int j = ; j < i; ++j) {
             for(int x = ; x < d; ++x) {
                 for(int y = ; y <= m; ++y) {
                     dp[i][(b[i] + x) % d][y] += dp[j][x][y - ];
                 }
             }
         }
     }
 }

 int main()
 {
     int t, n, q;
     scanf("%d", &t);
     for(int ca = ; ca <= t; ++ca) {
         scanf("%d %d", &n, &q);
         for(int i = ; i <= n; ++i) {
             scanf("%d", a + i);
         }
         printf("Case %d:\n", ca);
         while(q--) {
             int d, m;
             scanf("%d %d", &d, &m);
             for(int i = ; i <= n; ++i) {
                 b[i] = (a[i] % d + d) % d;
             }
             solve(n, m, d);
             LL ans = ;
             for(int i = ; i <= n; ++i) {
                 ans += dp[i][][m];
             }
             printf("%lld\n", ans);
         }
     }
     return ;
 }

优化后 0.012：

 #include <cstdio>
 #include <cstring>
 using namespace std;
 typedef long long LL;
 const int N = ;
 LL dp[][][]; //dp[i][j][k]表示前i个
 int a[N], b[N];

 void solve(int n, int m, int d) {
     memset(dp, , sizeof(dp));
     for(int i = ; i <= n; ++i) {
         dp[i][b[i]][] = ;
         for(int j = ; j < d; ++j) {
             for(int x = ; x <= m; ++x) {
                 dp[i][(b[i] + j) % d][x] += dp[i - ][j][x - ];
                 dp[i][j][x] += dp[i - ][j][x];
             }
         }
     }
 }

 int main()
 {
     int t, n, q;
     scanf("%d", &t);
     for(int ca = ; ca <= t; ++ca) {
         scanf("%d %d", &n, &q);
         for(int i = ; i <= n; ++i) {
             scanf("%d", a + i);
         }
         printf("Case %d:\n", ca);
         while(q--) {
             int d, m;
             scanf("%d %d", &d, &m);
             for(int i = ; i <= n; ++i) {
                 b[i] = (a[i] % d + d) % d;
             }
             solve(n, m, d);
             printf("%lld\n", dp[n][][m]);
         }
     }
     return ;
 }