HDU 5201 The Monkey King(容斥)
2015-04-09 21:03
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看了斌神的题解恍然大悟http://www.kuangbin.net/archives/bc36
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MOD = 1000000007;
const int N = 200005;
int t, n, m, f
, rf
;
int pow_mod(int x, int k) {
int ans = 1;
while (k) {
if (k&1) ans = (ll)ans * x % MOD;
x = (ll)x * x % MOD;
k >>= 1;
}
return ans;
}
int C(int n, int m) {
return (ll)f
* rf[n - m] % MOD * rf[m] % MOD;
}
int cal(int x, int n, int sum) {
if (n == 0) return sum == 0;
int ans = 0;
for (int i = 0; i * x <= sum && i <= n; i++) {
int tmp = (ll)C(n, i) * C(sum - i * x + n - 1, n - 1) % MOD;
if (i%2) ans = ((ans - tmp) % MOD + MOD) % MOD;
else ans = (ans + tmp) % MOD;
}
return ans;
}
int main() {
f[0] = 1;
for (int i = 1; i < N; i++)
f[i] = (ll)f[i - 1] * i % MOD;
rf[N - 1] = pow_mod(f[N - 1], MOD - 2);
for (int i = N - 2; i >= 0; i--)
rf[i] = (ll)rf[i + 1] * (i + 1) % MOD;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
int ans = 0;
for (int i = 1; i <= n; i++)
ans = (ans + cal(i, m - 1, n - i)) % MOD;
printf("%d\n", ans);
}
return 0;
}
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MOD = 1000000007;
const int N = 200005;
int t, n, m, f
, rf
;
int pow_mod(int x, int k) {
int ans = 1;
while (k) {
if (k&1) ans = (ll)ans * x % MOD;
x = (ll)x * x % MOD;
k >>= 1;
}
return ans;
}
int C(int n, int m) {
return (ll)f
* rf[n - m] % MOD * rf[m] % MOD;
}
int cal(int x, int n, int sum) {
if (n == 0) return sum == 0;
int ans = 0;
for (int i = 0; i * x <= sum && i <= n; i++) {
int tmp = (ll)C(n, i) * C(sum - i * x + n - 1, n - 1) % MOD;
if (i%2) ans = ((ans - tmp) % MOD + MOD) % MOD;
else ans = (ans + tmp) % MOD;
}
return ans;
}
int main() {
f[0] = 1;
for (int i = 1; i < N; i++)
f[i] = (ll)f[i - 1] * i % MOD;
rf[N - 1] = pow_mod(f[N - 1], MOD - 2);
for (int i = N - 2; i >= 0; i--)
rf[i] = (ll)rf[i + 1] * (i + 1) % MOD;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
int ans = 0;
for (int i = 1; i <= n; i++)
ans = (ans + cal(i, m - 1, n - i)) % MOD;
printf("%d\n", ans);
}
return 0;
}
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