写英文注释不是要“秀英文”，而是因为鄙人正在准备雅思，顺手练习
逆序对
题目描述
完整代码
#include<iostream>
using namespace std;
int num[500010]; // input numbers
int tmp[500010]; // sequence after merging left and right part
long long res;// Count of inversions
void merge(int left,int mid,int right){
int i=left,j=mid+1;
int idx=0;
while(i<=mid&&j<=right){
if(num[i]>num[j]){
tmp[idx++]=num[j++];
res+=mid-i+1;
}
else
tmp[idx++]=num[i++];
}
while(i<=mid)
tmp[idx++]=num[i++];
while(j<=right)
tmp[idx++]=num[j++];
}
void merge_sort(int left,int right){
if(left>=right){
return;
}
int mid=(left+right)/2;
merge_sort(left,mid);
merge_sort(mid+1,right);
merge(left,mid,right);
for(int i=left;i<=right;i++){
num[i]=tmp[i-left]; //num数组不是tmp数组翻转过来！
//这里是调整索引
}
}
int main(){
ios::sync_with_stdio(false);
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>num[i];
}
res=0;
merge_sort(0,n-1);
cout<<res<<endl;
}代码讲解
归并排序。在合并的过程发现逆序的时候，后面有多少个就说明有多少个逆序了。
for(int i=left;i<=right;i++){
num[i]=tmp[i-left]; //num数组不是tmp数组翻转过来！
//这里是调整索引
}注意这里的索引调整 i-left 是因为 tmp 数组是从 0 开始的，而 num 数组的相应段是从 left 开始的。
Max Sum
题目描述
完整代码
#include<iostream>
#include<cstring>
using namespace std;
int num[100010];
struct node{
int l,r,sum;
node(int x,int y,int z)
:l(x),r(y),sum(z){}//constructor initialization
};
node crossingSubArray(int num[],int mid,int low,int high){
node s3(0,0,0);
int sum=0;
int max_left=-10000;// max value for the left part
int max_right=-10000;// max value for the right part
// A very sophisticated algorithm. Since this array crosses both the left and right arrays, the middle must be crossed!
// For the left part, the sequence starts from the middle and goes to the low index. If there is an index that achieves a larger value, update the left index.
for(int i=mid;i>=low;i--){
sum+=num[i];
if(sum>=max_left){
max_left=sum;
s3.l=i;
}
}
sum=0;
// possess the right array in the same way
for(int i=mid+1;i<=high;i++){
sum+=num[i];
if(sum>max_right){
max_right=sum;
s3.r=i;
}
}
s3.sum=max_left+max_right;
return s3;
}
node maxSubArray(int num[],int left,int right){
if(left==right){//means that there is only one number
return node(left,right,num[left]);
}
int mid=(left+right)/2;
node s1= maxSubArray(num,left,mid);
node s2= maxSubArray(num,mid+1,right);
node s3= crossingSubArray(num,mid,left,right);
if(s1.sum>=s2.sum&&s1.sum>=s3.sum)
return s1;
else if(s2.sum>=s1.sum&&s2.sum>=s3.sum)
return s2;
else
return s3;
}
int main(){
int T,n;
ios::sync_with_stdio(false);
cin>>T;
int ncase=0;
while(T--){
cin>>n;
memset(num,0,sizeof(num));
for(int i=0;i<n;i++){
cin>>num[i];
}
node res=maxSubArray(num,0,n-1);
if(ncase){//equivalent to if(ncace!=0)
cout<<endl;
}
cout<<"Case "<<++ncase<<":\n";
cout<<res.sum<<" "<<res.l+1<<" "<<res.r+1<<endl;
}
}代码讲解
用分治法解决的思路是：
把数组对半分，使得和最大的左和右索引，要么包括了左半边，要么包括了右半边，要么一个在左边，一个在右边。所以在这三个中取最大值即可。如果只在某半边，因为最后会把那个块拆成单个，所以是可以直接通过比较比出来。如果是穿过了中间，注意要从中间往两边增（拆成先往左增再往右增）来比较怎么样才是最大的并且记下来此时的下标。
棋盘覆盖问题
题目描述
完整代码
#include<iostream>
#include<vector>
using namespace std;
int tile=1;// index of tile
void tileBoard(vector<vector<int> >& board,int x,int y,int specialX,int specialY,int size){
if(size==1) return;
int half=size/2;
int currentTile=tile++;
// top-left corner
// if the special block is in the top-left corner, solve the top-left corner first.
// if not, the top-left block among the four in the center can be determined as the place where the current tile should be placed
if (specialX < x + half && specialY < y + half) {
tileBoard(board, x, y, specialX, specialY, half);
} else {
board[x + half - 1][y + half - 1] = currentTile;
tileBoard(board, x, y, x + half - 1, y + half - 1, half);
}
if (specialX < x + half && specialY >= y + half) {
tileBoard(board, x, y + half, specialX, specialY, half);
} else {
board[x + half - 1][y + half] = currentTile;
tileBoard(board, x, y + half, x + half - 1, y + half, half);
}
// bottom-left corner
if (specialX >= x + half && specialY < y + half) { // in the bottom-left corner
tileBoard(board, x + half, y, specialX, specialY, half); // find the bottom-left corner
} else { // special block is not in the bottom-left corner
board[x + half][y + half - 1] = currentTile; // put current tile in BL corner
tileBoard(board, x + half, y, x + half, y + half - 1, half); // find the bottom-left corner
}
// bottom-right corner
if (specialX >= x + half && specialY >= y + half) {
tileBoard(board, x + half, y + half, specialX, specialY, half);
} else {
board[x + half][y + half] = currentTile;
tileBoard(board, x + half, y + half, x + half, y + half, half);
}
}
int main(){
int k,x,y;
cin>>k>>x>>y;
int size=(1<<k);
vector<vector<int> > board(size,vector<int>(size,0));
// a quick way to initialize the 2D vector to 0
// The number of dimensions in the vector corresponds to the number of levels of nesting
//NOTE:the coordinate offset
board[x-1][y-1]=0;
tileBoard(board,0,0,x-1,y-1,size);
for(const auto & row:board){
for(int i=0;i<row.size();i++){
cout<<row[i]<<(i==row.size()-1?'\n':' '); // A simple but useful judgement
}
}
}
//the board look like this
/*
* y 0 1 2
* x 0 TL TR
* 1 BL BR
* 2
*
*/代码讲解
这题的思路还有点绕。就是先把棋盘分成4大块。先看特殊方块在不在左上，如果在的话递归进入左上搜索。如果不在，说明左上最靠近中心（也就是左上大块最右下的小块）标记为当前的骨牌。其余几个同理。
a-Good String
题目描述
完整代码
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string str;
int solve(int left,int right,char c){
if(left==right){
if(c==str[left])
return 0;
else
return 1;
}
int mid =(left+right)/2;
int res1=0,res2=0;
for(int i=left;i<=mid;i++){
if(str[i]!=c) {
res1++;
}
}
for(int i=mid+1;i<=right;i++){
if(str[i]!=c) {
res2++;
}
}
return min(res1 + solve(mid + 1, right, c + 1), res2 + solve(left, mid, c + 1));
}
int main(){
int T,n;
cin>>T;
while(T--){
cin>>n>>str;
cout<<solve(0,n-1,'a')<<endl;
}
return 0;
}代码讲解
先拆分。用个solve函数返回当前字符串变多少个能成为good string。退出条件：如果最后只剩一个了，这个是指定的c，那就不用变，为0；否则，要变一个，为1。
然后尝试字符串前一半为a，根据题意，此时如果前一半中有不是a的，要变的次数+1。然后加上递归后一半中要变的。
同理再尝试字符串的后一半为a，根据题意，此时如果后一半中有不是a的，要变的次数+1。然后加上递归前一半中要变的。
最后，返回前后对比变的次数更少的那一个。