Light oj 1125

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 = 205;
LL dp[205][25][15]; //dp[i][j][k]表示包含i
int a[N], b[N];

void solve(int n, int m, int d) {
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= n; ++i) {
        dp[i][b[i]][1] = 1;
        for(int j = 1; j < i; ++j) {
            for(int x = 0; x < d; ++x) {
                for(int y = 2; y <= m; ++y) {
                    dp[i][(b[i] + x) % d][y] += dp[j][x][y - 1];
                }
            }
        }
    }
}

int main()
{
    int t, n, q;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d %d", &n, &q);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        printf("Case %d:
", ca);
        while(q--) {
            int d, m;
            scanf("%d %d", &d, &m);
            for(int i = 1; i <= n; ++i) {
                b[i] = (a[i] % d + d) % d;
            }
            solve(n, m, d);
            LL ans = 0;
            for(int i = 1; i <= n; ++i) {
                ans += dp[i][0][m];
            }
            printf("%lld
", ans);
        }
    }
    return 0;
}

优化后 0.012：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 205;
LL dp[205][25][15]; //dp[i][j][k]表示前i个
int a[N], b[N];

void solve(int n, int m, int d) {
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= n; ++i) {
        dp[i][b[i]][1] = 1;
        for(int j = 0; j < d; ++j) {
            for(int x = 1; x <= m; ++x) {
                dp[i][(b[i] + j) % d][x] += dp[i - 1][j][x - 1];
                dp[i][j][x] += dp[i - 1][j][x];
            }
        }
    }
}

int main()
{
    int t, n, q;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d %d", &n, &q);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        printf("Case %d:
", ca);
        while(q--) {
            int d, m;
            scanf("%d %d", &d, &m);
            for(int i = 1; i <= n; ++i) {
                b[i] = (a[i] % d + d) % d;
            }
            solve(n, m, d);
            printf("%lld
", dp[n][0][m]);
        }
    }
    return 0;
}