B. So Easy
Mr. G invents a new game whose rules are given as follows.
Firstly, he has a n \times nn×n matrix, all elements of which are 00 initially. Then, he follows up with some operations: in each time he chooses a row or a column, and adds an arbitrary positive integer to all the elements in the selected row or column. When all operations have been finished, he hides an element in the matrix and the element is modified to -1−1.
Now given the final matrix, you are asked to find out what the hidden element should be before the very last hiding operation.
Input
The first line contains a single integer n(2≤n≤1000).
Next nn lines represent the matrix after the operations. Each element in the matrix satisfies −1≤a i,j≤1000000, and exactly one element is -1.
Output
Output a single integer, the hidden element.
樣例輸入
3
1 2 1
0 -1 0
0 1 0
樣例輸出
1
題意
n×n的數值矩陣，初始數值全爲 0，你可以對每一行或者每一列同時加某一正整數，得到最終矩陣，問你某一點的數值（輸入是-1的點）
思路
- 找到 -1 點的位置(ai,aj)，記錄，並把這個點重新賦值爲 0
- 先進行行操作，找每一行的最小值，再每個數都減去這個最小值，注意的是最小值尋找時，要避開(ai,aj)點
- 同理進行列操作
- 輸出 -a[ai][aj]
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
#include<utility>
#include<map>
#define ll long long
#define ld long double
#define ull unsigned long long
using namespace std;
const int INF = 0x3f3f3f3f3f;
const double eps = 1e-6;
const int maxn = 1010;
int a[maxn][maxn];
int main(void)
{
int n;
int ai,aj;
scanf("%d",&n);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
scanf("%d",&a[i][j]);
if(a[i][j]==-1){
ai = i;aj = j;
a[i][j] = 0;
}
}
}
int _min;
for(int i=1;i<=n;i++){
_min = INF;
for(int j=1;j<=n;j++){
if(i!=ai||j!=aj){
_min = min(_min,a[i][j]);
}
}
for(int j=1;j<=n;j++){
a[i][j] -= _min;
}
}
for(int j=1;j<=n;j++){
_min = INF;
for(int i=1;i<=n;i++){
if(i!=ai||j!=aj) _min = min(_min,a[i][j]);
}
for(int i=1;i<=n;i++){
a[i][j] -= _min;
}
}
printf("%d\n",-a[ai][aj]);
return 0;
}
I Base 62
As we already know, base64 is a common binary-to-text encoding scheme. Here we define a special series of positional systems that represent numbers using a base (a.k.a. radix) of 2 to 62. The symbols ‘0’ – ‘9’ represent zero to nine, and ‘A’ – ‘Z’ represent ten to thirty-five, and ‘a’ – ‘z’ represent thirty-six to sixty-one. Now you need to convert some integer z in base xx into base y.
Input
The input contains three integers x, y (2≤x,y≤62) and z (0≤z<x120), where the integer z is given in base x.
Output
Output the integer zz in base yy.
樣例輸入
16 2 FB
樣例輸出
11111011
題意
把一個 x 進制數 z 轉化成 y 進制數，輸出注意：0 ~ 9對應’0’ ~ 9’，10 ~ 35對應 ‘A’ ~ ‘Z’，36 ~ 61對應 ‘a’ ~ ‘z’。
思路
進制轉換的模板題目：
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
#include<utility>
#include<map>
#define ll long long
#define ld long double
#define ull unsigned long long
using namespace std;
const int INF = 0x3f3f3f3f3f;
const double eps = 1e-6;
const int maxn = 10000010;
char str[maxn],another[maxn];
int ten[maxn];
int switchToTen(int m)
{
int i,j,len,k,c;
len = strlen(str);
k = 1;
memset(ten,0,sizeof(ten));
for(int i = 0;i<len;i++){
for(int j=0;j<k;j++){
ten[j] *= m;
}
if(str[i]>='0'&&str[i]<='9'){
ten[0] += str[i]-'0';
}
else if(str[i]>='A'&&str[i]<='Z'){
ten[0] += str[i]-'A'+10;
}
else if(str[i]>='a'&&str[i]<='z'){
ten[0] += str[i]-'a'+36;
}
for(int j=c=0;j<k;j++){
ten[j] += c;
if(ten[j]>=10){
c = ten[j] / 10;
ten[j] %= 10;
}
else c = 0;
}
while(c){
ten[k++] = c % 10;
c /= 10;
}
}
int temp;
for(int i=0,j=k-1;i<j;i++,j--){
temp = ten[i];
ten[i] = ten[j];
ten[j] = temp;
}
return k;
}
void switchToAnother(int k,int n)
{
int sum,r,t,d;
sum = 1;
r = 0;
memset(another,0,sizeof(another));
while(sum){
sum = 0;
for(int i=0;i<k;i++){
d = ten[i] / n;
sum += d;
if(i==k-1){
t = ten[i] % n;
if(t>=0&&t<=9){
another[r] = t+'0';
}
else if(t>=10&&t<=35){
another[r] = t-10+'A';
}
else another[r] = t-36+'a';
r++;
}
else{
ten[i+1] += ten[i]%n * 10;
}
ten[i] = d;
}
}
for(int i=r-1;i>=0;i--){
printf("%c",another[i]);
}
printf("\n");
}
int main()
{
int m,n,k;
cin>>m>>n>>str;
k = switchToTen(m);
switchToAnother(k,n);
return 0;
}
G. Pot!!
樣例輸入：
5 6
MULTIPLY 3 5 2
MULTIPLY 2 5 3
MAX 1 5
MULTIPLY 1 4 2
MULTIPLY 2 5 5
MAX 3 5
樣例輸出：
ANSWER 1
ANSWER 2
題意
給以一段 1~n 的區間(tree)，他們的初始值爲 1，有兩個操作：
- MULTIPLY l r x 操作：讓[l,r] 內的每個數乘以 x (2<=x<=10)
- MAX l r 操作：輸出[l,r]的最大 m，tree[i] % pm = 0 && tree[i] % pm+1 != 0，i∈[l,r]，p是任意素數。
思路
區間修改和查詢問題很明顯是線段樹問題，這裏的MAX操作像是一個質因分解操作。
打素數表，然後對每一個數質因分解？？
我們想一下，x的取值範圍[2,10]，那麼這些區間裏的質因子一定是<=10的，那麼就是 2 3 5 7 這四個數了
那麼就線段樹維護 2 3 5 7 這四個值得 mx，然後MAX是查詢最大mx就行了。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
#include<utility>
#include<map>
#define ll long long
#define ld long double
#define ull unsigned long long
using namespace std;
const int INF = 0x3f3f3f3f3f;
const double eps = 1e-6;
const int maxn = 100010;
int p[maxn];
int n,m;
char cmd[10];
int x,y,z;
struct node{
int l,r,mx;
int mx2,la2;
int mx3,la3;
int mx5,la5;
int mx7,la7;
}a[maxn<<2];
void update(int k)
{
a[k].mx2 = max(a[k<<1].mx2,a[k<<1|1].mx2);
a[k].mx3 = max(a[k<<1].mx3,a[k<<1|1].mx3);
a[k].mx5 = max(a[k<<1].mx5,a[k<<1|1].mx5);
a[k].mx7 = max(a[k<<1].mx7,a[k<<1|1].mx7);
a[k].mx = max(a[k<<1].mx,a[k<<1|1].mx);
}
void pushdown(int k)
{
if(a[k].la2){
a[k<<1].la2 += a[k].la2;
a[k<<1|1].la2 += a[k].la2;
a[k<<1].mx2 += a[k].la2;
a[k<<1|1].mx2 += a[k].la2;
a[k].la2 = 0;
}
if(a[k].la3){
a[k<<1].la3 += a[k].la3;
a[k<<1|1].la3 += a[k].la3;
a[k<<1].mx3 += a[k].la3;
a[k<<1|1].mx3 += a[k].la3;
a[k].la3 = 0;
}
if(a[k].la5){
a[k<<1].la5 += a[k].la5;
a[k<<1|1].la5 += a[k].la5;
a[k<<1].mx5 += a[k].la5;
a[k<<1|1].mx5 += a[k].la5;
a[k].la5 = 0;
}
if(a[k].la7){
a[k<<1].la7 += a[k].la7;
a[k<<1|1].la7 += a[k].la7;
a[k<<1].mx7 += a[k].la7;
a[k<<1|1].mx7 += a[k].la7;
a[k].la7 = 0;
}
a[k<<1].mx = max(max(a[k<<1].mx2,a[k<<1].mx3),max(a[k<<1].mx5,a[k<<1].mx7));
a[k<<1|1].mx = max(max(a[k<<1|1].mx2,a[k<<1|1].mx3),max(a[k<<1|1].mx5,a[k<<1|1].mx7));
}
void build(int k,int l,int r)
{
a[k].l = l;a[k].r = r;
a[k].la2 = a[k].la3 = 0;
a[k].la5 = a[k].la7 = 0;
a[k].mx2 = a[k].mx3 = 0;
a[k].mx5 = a[k].mx7 = 0;
if(l==r) return;
int mid=(l+r)>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
update(k);
}
void changeSegment(int k,int l,int r,int x2,int x3,int x5,int x7)
{
if(a[k].l>=l&&a[k].r<=r){
a[k].la2 += x2;a[k].la3 += x3;
a[k].la5 += x5;a[k].la7 += x7;
a[k].mx2 += x2;a[k].mx3 += x3;
a[k].mx5 += x5;a[k].mx7 += x7;
a[k].mx = max(max(a[k].mx2,a[k].mx3),max(a[k].mx5,a[k].mx7));
return;
}
pushdown(k);
int mid=(a[k].l+a[k].r)>>1;
if(r>mid) changeSegment(k<<1|1,l,r,x2,x3,x5,x7);
if(l<=mid) changeSegment(k<<1,l,r,x2,x3,x5,x7);
update(k);
}
int query(int k,int l,int r)
{
if(a[k].l>=l&&a[k].r<=r) return a[k].mx;
pushdown(k);
int x = 0;
int mid=(a[k].l+a[k].r)>>1;
if(r>mid) x = max(x,query(k<<1|1,l,r));
if(l<=mid) x = max(x,query(k<<1,l,r));
return x;
}
int main(void)
{
scanf("%d%d",&n,&m);
build(1,1,n);
for(int i=1;i<=m;i++){
scanf("%s",cmd);
if(cmd[1]=='U'){
scanf("%d%d%d",&x,&y,&z);
int x2,x3,x5,x7;
x2 = x3 = x5 = x7 = 0;
if(z==2) x2++;
else if(z==3) x3++;
else if(z==4) x2 += 2;
else if(z==5) x5++;
else if(z==6) x2++,x3++;
else if(z==7) x7++;
else if(z==8) x2 += 3;
else if(z==9) x3 += 2;
else if(z==10) x2++,x5++;
changeSegment(1,x,y,x2,x3,x5,x7);
}
else{
scanf("%d%d",&x,&y);
printf("ANSWER %d\n",query(1,x,y));
}
}
return 0;
}