CodeForces 314C Sereja and Subsequences ：找出序列种所有不下降子序列，求出所有子序列乘积和 ：技巧/树状数组

树状数组数组里面记录的值是当前以a[i]结尾的乘积和=

比如当前到i的a[i],查询1-a[i]的所有值和，乘以(a[i]+1)就是当前以a[i]为结尾的乘积和=

这样出现了重复计算的情况，只要减去上一次a[i]结尾的乘积和即可

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL long long
const LL MOD=1e9+7;
LL m,dp[1000005],a[100005];
LL sum(LL x)
{
  LL ret=0;
  while (x>0){
    ret=(ret+dp[x])%MOD;
    x-=x&-x;
  }
  return ret;
}
void add(LL x,LL y)
{
  while (x<=m){
    dp[x]=(dp[x]+y)%MOD;
    x+=x&-x;
  }
}
int main()
{
  LL n,i,ans,t1,t2,val;
  m=ans=0;
  scanf("%I64d",&n);
  for (i=1;i<=n;i++){
       scanf("%I64d",&a[i]);
    m=max(m,a[i]);
  }
  memset(dp,0,sizeof(dp));
  for (i=1;i<=n;i++)
  {
    t1=sum(a[i]); t2=sum(a[i]-1);
    val=((t1+1)*a[i]-(t1-t2+MOD)%MOD)%MOD;
    add(a[i],val);
    ans=(ans+val)%MOD;
  }
  printf("%I64d\n",ans);
  return 0;
}

题目链接：http://codeforces.com/problemset/problem/314/C