Find the answer
Description
Given a sequence of n integers called W and an integer m. For each i (1 <= i <= n), you can choose some elements Wk (1 <= k < i), and change them to zero to make ∑ij=1Wj<=m. So what's the minimum number of chosen elements to meet the requirements above?.
Input
The first line contains an integer Q --- the number of test cases.
For each test case:
The first line contains two integers n and m --- n represents the number of elemens in sequence W and m is as described above.
The second line contains n integers, which means the sequence W.
1 <= Q <= 15
1 <= n <= 2*105
1 <= m <= 109
For each i, 1 <= Wi <= m
output
For each test case, you should output n integers in one line: i-th integer means the minimum number of chosen elements Wk (1 <= k < i), and change them to zero to make ∑ij=1Wj<=m.
Examples
Input
Output
0 0 0 0 0 2 3
0 1 1 2 3
正确解法：
对于每一个i来说，使前面尽可能多的a[j]+a[i] <=m （j<i）
也就是说对前面 i-1 个数排序，找最多的个数。使他们加起来 <= m-a[i]
我们用权值线段树来搞。
权值线段树什么意思呢？就是类似一个桶排序的东西，树的节点统计个数。
当 l==r==a[l] tree[rt]++;
树的范围就是a[i]，可题目中a[i]<=1e9，所以我们要离散化。
对于每个i来说，我们找最多的小数个数 res=m-a[i];
我们查询 1-cnt 中 1-mid 数的总和，若左儿子的总和小于res 的话，那就左儿子的个数+查询右儿子。
else 查询左儿子。
查询过a[i]后，我们把a[i]放进树中。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<string> 6 #include<map> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<cmath> 11 #include<cstdlib> 12 #include<stack> 13 #define de printf(" debug: ") 14 #define End printf(" end ") 15 #define fi first 16 #define se second 17 #define P pair< int, int > 18 #define PII pair< pair<int, int> ,int> 19 #define INF 0x3f3f3f3f 20 using namespace std; 21 typedef long long ll; 22 const int mod1=1e5+7; 23 const int N=2e5+100; 24 const int mod2=1e9+7; 25 const ll mod3=1e12+7; 26 int T,n,m; 27 int a[N],b[N],cnt=0,c[N]; 28 ll tree[N*4],totsum[N*4]; 29 void push_up(int rt) 30 { 31 tree[rt]=tree[rt<<1]+tree[rt<<1|1]; 32 totsum[rt]=totsum[rt<<1]+totsum[rt<<1|1]; 33 } 34 void update(int x,int l,int r,int rt) 35 { 36 if(l==r) 37 { 38 tree[rt]++; 39 totsum[rt]+=b[x]; 40 return ; 41 } 42 int mid=l+r>>1; 43 if(x<=mid) update(x,l,mid,rt<<1); 44 else update(x,mid+1,r,rt<<1|1); 45 push_up(rt); 46 } 47 int query(int l,int r,int rt,int res) 48 { 49 if(l==r) 50 { 51 return res/b[l]; 52 } 53 ll sum=totsum[rt<<1]; 54 int mid=l+r>>1; 55 if(sum>=res) 56 return query(l,mid,rt<<1,res); 57 else 58 return tree[rt<<1]+query(mid+1,r,rt<<1|1,res-sum); 59 } 60 int main() 61 { 62 scanf("%d",&T); 63 while(T--) 64 { 65 scanf("%d%d",&n,&m); 66 memset(tree,0,sizeof(tree)); 67 memset(totsum,0,sizeof(totsum)); 68 for(int i=1;i<=n;i++) 69 { 70 scanf("%d",&a[i]); 71 b[i]=a[i]; 72 } 73 sort(b+1,b+n+1); 74 cnt=unique(b+1,b+n+1)-b-1; 75 for(int i=1;i<=n;i++) 76 c[i]=lower_bound(b+1,b+cnt+1,a[i])-b; 77 ll tot=0; 78 for(int i=1;i<=n;i++) 79 { 80 tot+=a[i]; 81 if(tot<=m) 82 printf("0 "); 83 else 84 { 85 int ans=query(1,cnt,1,m-a[i]); 86 printf("%d ",i-ans-1); 87 } 88 update(c[i],1,cnt,1); 89 } 90 printf(" "); 91 } 92 93 94 return 0; 95 }
PS: CF中有一道很像的题目，就是数据范围小了一点，我们就可以真的用桶排序来写了。