标签：
HDU - 2227
Description How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1,
2, 3}.
Input The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, ...., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.
Output For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.
Sample Input
Sample Output
题意:给予一个序列求解所构成的不下降子序列的个数
通过离散化，将0->2^32分散到0->1e5之中
到达j处的不下降子序列的个数dp[j] = sum{dp[i]} + 1 {i < j && S[i] <= S[j] && i > 0}; 状态方程推导:
假设dp[j]为S[j]为结尾的序列的个数
那么if(S[j] < S[i])那么 dp[i] += dp[j] 为什么呢，举个简单的例子 如果以S[j]为结尾的序列是一下: 最原始的序列:S = { 1, 5, 3, 6, 7, 5, 2, 1, 8, 9}; 如果j = 4的话 那么以S[j]为结尾的不下降子序列如下: 6 1 6 5 6 3 6 1 3 6 1 5 6 那么如果我们取i = 5时 S[i] >= S[j]可以得到 6 7 1 6 7 5 6 7 3 6 7 1 3 6 7 1 5 6 7 就是dp[i] += dp[j]如此将所有S[j] <= S[i]都加起来 然后加上本身7加上那么状态转移方程极为dp[i] = sum{dp[j]} + 1 如此计算sum{dp[j],i < j && S[i] <= S[j] && i > 0}便是我们的难题了
而求和问题解决的话，可以基本能分为三类：
一：树状数组
二：线段树
三：前缀和
此处则是用了树状数组求解满足条件的数据和
 #include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 5;
const LL mod = 1e9 + 7;
int N, A[MAXN], S[MAXN];
LL C[MAXN];
int Crete(LL x) {
int lb = 0,ub = N;
while(ub - lb > 1) {
int mid = (lb + ub) >> 1;
if(A[mid] >= x) ub = mid;
else lb = mid;
}
return ub;
}
int lowbit(int x) {
return x & (-x);
}
void add(int x,LL c) {
while(x <= N) {
C[x] += c;
C[x] %= mod;
x += lowbit(x);
}
}
int sum(int x) {
LL ret = 0;
while(x > 0) {
ret += C[x];
ret %= mod;
x -= lowbit(x);
}
return ret;
}
int main() {
while(~ scanf("%d", &N)) {
for(int i = 1; i <= N; i ++) {
scanf("%d", &S[i]);
A[i] = S[i];
}
sort(A + 1, A + N + 1);
LL ans = 0;
memset(C, 0, sizeof(C));
for(int i = 1; i <= N; i ++) {
int ret = lower_bound(A + 1, A + N + 1, S[i]) - A;
//int ret = Crete(S[i]);
int res = sum(ret) + 1;
ans += res;
ans %= mod;
add(ret, res);
}
printf("%I64d\n", ans % mod);
}
return 0;
}
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HDU - 2227 Find the nondecreasing subsequences (树状数组 + 子序列 + 离散化)
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原文地址：http://blog.csdn.net/qq_18661257/article/details/47399287
