最后一题完全不会做，暴力都不会暴力，骗了30%。。。。。。。
A. 没啥难度，但是如果开vector[50000]这种C++会内存超限，非常恶心，所以直接用链表存了
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* @param m int整型
* @param a ListNode类 指向彩带的起点，val表示当前节点的val，next指向下一个节点
* @return ListNode类vector
*/
/*vector<int>vec[500005];
vector<ListNode*> solve(int m, ListNode* a) {
// write code here
while(a!=NULL){
vec[a->val%m].push_back(a->val);
a = a->next;
}
vector<ListNode*>ans;
for(int i=0;i<m;i++){
int size = vec[i].size();
if(size == 0)
ans.push_back(NULL);
else{
ListNode* head = new ListNode(-1);
ListNode* tmp = head;
for(int j=0;j<size;j++){
ListNode* p = new ListNode(vec[i][j]);
tmp->next = p;
tmp = p;
}
ans.push_back(head->next);
}
}
return ans;
}*/
vector<ListNode*>ans;
map<int,ListNode*>t;
vector<ListNode*> solve(int m, ListNode* a) {
// write code here
for(int i=0;i<m;i++){
ListNode* h = new ListNode(-1);
ans.push_back(h);
t[i] = h;
}
while(a!=NULL){
int dir = a->val%m;
ListNode* n = new ListNode(a->val);
ListNode* be = t[dir];
be->next = n;
t[dir] = n;
a = a->next;
}
for(int i=0;i<m;i++){
ListNode* be = ans[i];
ans[i] = be->next;
}
return ans;
}
}; B.N方过不了，标记一下到当前位置总共加了多少value即可，复杂度N
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t;
ll num[1005];
const ll mod = 1e9+7;
int main()
{
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&num[i]);
sort(num+1,num+1+n);
ll ans = 0;
ll sum = 0;
for(int i=n;i>=1;i--)
{
num[i] = (num[i] + sum)%mod;
ans = (ans + num[i])%mod;
sum = (sum + num[i])%mod;
}
printf("%lld\n",ans);
}
return 0;
} #include <bits/stdc++.h>
using namespace std;
#define ll long long
int t;
int even[1005];
int odd[1005];
int main()
{
scanf("%d",&t);
while(t--){
int n,m;
scanf("%d%d",&n,&m);
int oc = 0;
int ec = 0;
for(int i=1;i<=n;i++)
{
int tmp;
scanf("%d",&tmp);
if(tmp%2==1)
odd[++oc] = tmp;
else
even[++ec] = tmp;
}
int ans = n;
sort(even+1,even+1+ec);
sort(odd+1,odd+1+oc);
int l = 1,r=oc;
while(l<r){
while(l<r){
if(odd[l]+odd[r]<=m)
{
ans--;
r--;
break;
}
r--;
}
l++;
}
l=1,r=ec;
while(l<r){
while(l<r){
if(even[l]+even[r]<=m)
{
ans--;
r--;
break;
}
r--;
}
l++;
}
printf("%d\n",ans);
}
return 0;
} #include <bits/stdc++.h>
using namespace std;
#define ll long long
int n,k;
int main()
{
scanf("%d%d",&n,&k);
string str;
cin>>str;
string ans = "";
int len = str.length();
int now = 0;
/*i 后面 还有 k-i
至少要留出k-i个字符
len - 1
len - k + i*/
for(int i=1;i<=k;i++){
char large = 'a' - 1;
int dir;
for(int j=now;j<len-k+i;j++){
if(large<str[j]){
large = str[j];
dir = j;
}
}
ans.push_back(large);
now = dir+1;
}
cout<<ans<<endl;
return 0;
} 
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