hdu 4658 Integer Partition（公式）

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题目链接：hdu 4658 Integer Partition

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;const int mod = 1e9 + 7;int f[maxn];void init () {    f[0] = 1;    for (int i = 1; i <= maxn; i++) {        int bit = 1;        for (int j = 1; j <= i; j++) {            int v1 = i - j * (3 * j - 1) / 2;            int v2 = i - j * (3 * j + 1) / 2;            if (v1 < 0 && v2 < 0) break;            if (v1 >= 0)                f[i] = (f[i] + bit * f[v1]) % mod;            if (v2 >= 0)                f[i] = (f[i] + bit * f[v2]) % mod;            bit *= -1;        }        f[i] = (f[i] + mod) % mod;    }}int solve(int n, int k) {    int ret = f[n], sig = -1;    for (int i = 1; ; i++) {        int v1 = k * i * (3 * i - 1) / 2;        int v2 = k * i * (3 * i + 1) / 2;        if (v1 > n && v2 > n) break;        if (v1 <= n) ret = (ret + sig * f[n-v1]) % mod;        if (v2 <= n) ret = (ret + sig * f[n-v2]) % mod;        sig *= -1;    }    return (ret + mod) % mod;}int main () {    init();    int cas, n, k;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d", &n, &k);        printf("%d\n", solve(n, k));    }    return 0;}

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