CF - 314C - Sereja and Subsequences(树状数组+dp)
2013-08-15 21:34
489 查看
题意:一个由n个数a1, a2, ..., an组成的序列,对于这个序列的任何一个不同的不减子序列,x1, x2, ..., xr,存在y = {y1, y2, ..., yr},使得y1 <= x1, y2 <= x2, ..., yr <= xr,求y的总个数(1 <= n <= 10^5, 1 <= ai <= 10^6)。
题目链接:http://codeforces.com/problemset/problem/314/C
——>>设d[a]表示以数a结尾的子序列的y的个数,则
状态转移方程为:d[a] = sum(a) * a + a。
样例:1 2 2
对于1,d[1] = 1,
对于第1个2,d[2] = 4,
对于第2个2,(暂不赋值d[2]) temp = (d[1] + d[2]) * 2 + 2 = 12,这时,数状数组里的C[2]应加上的是temp - d[2](d[2]指上一个d[2] == 4),因为以2为结尾的y已有4个,所以temp中的12个中有4个是重复的,再更新d[2] = 12。
题目链接:http://codeforces.com/problemset/problem/314/C
——>>设d[a]表示以数a结尾的子序列的y的个数,则
状态转移方程为:d[a] = sum(a) * a + a。
样例:1 2 2
对于1,d[1] = 1,
对于第1个2,d[2] = 4,
对于第2个2,(暂不赋值d[2]) temp = (d[1] + d[2]) * 2 + 2 = 12,这时,数状数组里的C[2]应加上的是temp - d[2](d[2]指上一个d[2] == 4),因为以2为结尾的y已有4个,所以temp中的12个中有4个是重复的,再更新d[2] = 12。
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int mod = 1000000000 + 7; const int maxn = 1000000 + 10; const int N = 1000000 + 1; int n, d[maxn], C[maxn]; int lowerbit(int x){ return x & (-x); } void add(int x, int v){ v = (v % mod + mod) % mod; while(x <= N){ C[x] = (C[x] + v) % mod; x += lowerbit(x); } } int sum(int x){ int ret = 0; while(x){ ret = (ret + C[x]) % mod; x -= lowerbit(x); } return ret; } void init(){ memset(d, 0, sizeof(d)); memset(C, 0, sizeof(C)); } void solve(){ int a, i; for(i = 1; i <= n; i++){ scanf("%d", &a); int temp = ((long long)sum(a) * a + a) % mod; add(a, temp - d[a]); d[a] = temp; } printf("%d\n", sum(N)); } int main() { while(scanf("%d", &n) == 1){ init(); solve(); } return 0; }
相关文章推荐
- 【codeforces 314C】Sereja and Subsequences
- CF380A Sereja and Prefixes
- CF_314D_Sereja and Periods
- cf-381A-Sereja and Dima
- cf B. Sereja and Suffixes
- CF_315A_Sereja and Bottles
- CF 314B: Sereja and Periods
- CF-Sereja and Algorithm
- CF_315B_Sereja and Array
- CF314-C Sereja and Subsequences
- cf C. Sereja and Algorithm
- CF 361 D. Friends and Subsequences (RMQ+二分查找)
- 【CF314C】Sereja and Subsequences(DP,树状数组)
- CF_315C_Sereja and Contest
- Codeforces 315E Sereja and Subsequences【思维+Dp】
- cf Sereja and Array
- B. Sereja and Suffixes(cf)
- CF 315B - Sereja and Array(树状数组)
- 【CF】223 Div.1 C Sereja and Brackets
- CF 380C - Sereja and Brackets(线段树)