题目链接:HDU4602
【题意】给出n,k(<=10^9),在用1~n的数字的所有排列相加==n中,数字k出现的次数;
【分析】首先很容易发现任何n出现次数都是由1,2,5,12…这样一个规律倒着出现,例如n = 4时,1出现12次,2出现5次,3出现2次,4出现1次…这样类推,用cnt[i]记录这个1,2,5,12…序列,则n,k的答案就是cnt[n-k+1];当k>n是要特判为0
然后就是怎么快速求出这个序列了,通过暴力打表发现一个递推式:cnt[i]=sum[i-1]+p[i-1];其中cnt[i]表示第i个数字出现的次数,sum[i]是cnt[1]~cnt[i]的和,p[i-1]是2^(i-1);但是n有10^9,显然不能用递推式求和存;然后想到了用矩阵乘法求;为了简单把原式转换一下变成sum[i]=2*sum[i-1]+p[i-1];这样cnt[i]=sum[i]-sum[i-1];sum[1] = 1;然后只要构造出一个矩阵
X={2,0
1,2};
在构造一个矩阵A={sum[i-1],p[i-1]}
则A*X = {2*sum[i-1]+p[i-1], 2*p[i-1]} == {sum[i],p[i]};
这样A* X^n-1次就是{sum[n],p[n]};sum[n]算出来,sum[n-1]再算一下,答案就出来了;
【AC CODE】265ms
#include <cstdio> #include <cstring> #include <cctype> #include <cmath> #include <map> //#include <unordered_map> #include <queue> #include <stack> #include <vector> #include <string> #include <algorithm> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i = a; i <= n; i++) #define per(i,n,a) for(int i = n; i >= a; i--) #define clc(a,b) memset(a,b,sizeof(a)) const int INF = 0x3f3f3f3f, MOD = 1000000000+7; /* sum[i] = 2*sum[i-1]+p[i-1]&&p[i]=2*p[i-1] sum[1] = 1 p[1] = 1*/ struct MATRIX{ LL num[2][2]; MATRIX(LL a1, LL a2, LL b1, LL b2) { num[0][0] = a1, num[1][0] = a2; num[0][1] = b1, num[1][1] = b2; } }; MATRIX x = MATRIX(2,1,0,2); MATRIX mul(const MATRIX &a, const MATRIX &b) { MATRIX ans = MATRIX(0,0,0,0); rep(i,0,2) { rep(j,0,2) { rep(k,0,2) ans.num[i][j] = (ans.num[i][j]+a.num[i][k]*b.num[k][j]%MOD)%MOD; } } return ans; } MATRIX pow_mod(MATRIX x, int n) { MATRIX ans = MATRIX(1,0,0,1); while(n) { if(n&1) ans = mul(ans,x); x = mul(x,x); n >>= 1; } return ans; } int main() { #ifdef SHY freopen("e:\\1.txt", "r", stdin); #endif int t; scanf("%d%*c", &t); while(t--) { int n,k; scanf("%d %d%*c", &n, &k); if(k >= n) { if(k == n) puts("1"); else puts("0"); continue; } n = n-k+1; MATRIX p = pow_mod(x,n-1); LL ans = (p.num[0][0]+p.num[1][0])%MOD; p = pow_mod(x,n-2); ans = (ans-(p.num[0][0]+p.num[1][0])%MOD+MOD)%MOD; printf("%I64d\n", ans); } return 0; }
【暴力代码】
#include <cstdio> #include <cstring> #include <cctype> #include <cmath> #include <map> //#include <unordered_map> #include <queue> #include <stack> #include <vector> #include <string> #include <algorithm> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i = a; i <= n; i++) #define per(i,n,a) for(int i = n; i >= a; i--) #define clc(a,b) memset(a,b,sizeof(a)) const int INF = 0x3f3f3f3f, MAXN = 1000; int cnt[MAXN], n, num[MAXN]; void dfs(int sum, int p) { if(sum == n) { repe(i,1,p) cnt[num[i]]++; return; } repe(i,1,n-sum) { num[p+1] = i; dfs(sum+i,p+1); } } int main() { #ifdef SHY freopen("e:\\1.txt", "r", stdin); #endif int t; n = 20; clc(cnt,0); dfs(0,0); int sum = 0; printf("num cnt\t\tpow\tsum\tsum-pow\t\tcnt-pow\n"); repe(i,1,n) { sum += cnt[n-i+1]; printf("%2.d: %-10.d %-10.0lf %-10d %-10.0lf %-10.0lf %-10.0lf\n", i,cnt[n-i+1], pow(2,i-1),sum,sum-pow(2,i-1),cnt[n-i+1]-pow(2,i-1),sum-cnt[n-i+1]+pow(2,i-2)); //printf("%d-%d: %d\n", i,i-1,cnt[n-i+1]-cnt[n-i+2]); } return 0; }