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核桃的数量
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
int gcd(int a, int b){
return (b == 0)? a: gcd(b,a%b);
}
int main(){
int a, b, c, d;
cin>>a>>b>>c;
int A = a/gcd(a,b)*b, B = a/gcd(a,c)*c;
cout<<A/gcd(A,B)*B<<endl;
return 0;
}
合根植物
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn = (1000+10) * (1000+10);
int fa[maxn];
int n, m, k, N;
int a, b;
int findfa(int x){
while(x != fa[x]) x = fa[x];
return x;
}
int main(){
cin>>n>>m>>k;
N = n*m;
for(int i = 1; i <= N; i++) fa[i] = i;
while(k--){
cin>>a>>b;
int A = findfa(a), B = findfa(b);
if(A != B){
fa[A] = B;
N--;
}
}
cout<<N<<endl;
return 0;
}
/*
sample input
5 4
16
2 3
1 5
5 9
4 8
7 8
9 10
10 11
11 12
10 14
12 16
14 18
17 18
15 19
19 20
9 13
13 17
sample output
5
*/
小数第n位
// 注意弯道超车
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long ll;
ll a, b, n, t, k = 3, cnt = 0;
int main(){
cin>>a>>b>>n;
a = a%b;
while(n-10 > 3){ //speed up
a = (a * 10000000000) % b;
n -= 10;
if(a == 0) {
cout<<"000\n";
return 0;
}
}
while(k){
a = a*10;
t = a / b;
a = a % b;
if(++cnt >= n){
cout<<t;
k--;
}
}
return 0;
}
/*
sample input
1 8 1
1 8 3
282866 999000 6
sample output
125
500
914
*/
回文数字
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
int ok = 0;
int n;
void solve0(){
for(int x = 1; x < 10; x++){
for(int y = 0; y < 10; y++) {
int z = n-2*x-2*y;
if( z>=0 && z<=9){
cout<<x<<y<<(n-2*x-2*y)<<y<<x<<endl;
ok = 1;
}
}
}
}
void solve1(){
for(int x = 1; x < 10; x++){
for(int y = 0; y < 10; y++) {
int z = n-2*x-2*y;
if( z%2==0 && z>=0 && z<=18){
cout<<x<<y<<z/2<<z/2<<y<<x<<endl;
ok = 1;
}
}
}
}
int main(){
cin>>n;
solve0(); solve1();
if(!ok) cout<<"-1\n";
return 0;
}
/*
sample input
40
sample output
99899
499994
589985
598895
679976
688886
697796
769967
778877
787787
796697
859958
868868
877778
886688
895598
949949
958859
967769
976679
985589
994499
*/
数字游戏
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long ll;
int main()
{
ll n, k, T;
ll sum = 0;
cin>>n>>k>>T;
for(ll i = 0; i < T ; i++){
ll t = 1+i*n;
if(t%2==0) sum+= (1+( ((t/2)%k) *((t-1)%k) )%k )%k;
else sum+= (1+( (t%k) *( ((t-1)/2)%k) )%k )%k;
}
cout<<sum<<endl;
return 0;
}
/*
sample input
999999
999999
999999
sample output
999999
*/
九宫幻方
我们先将所有结果跑出放到一个数组中
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
char A[8][10]=
{"276951438",
"294753618",
"438951276",
"492357816",
"618753294",
"672159834",
"816357492",
"834159672"};
int main(){
int a[9], cnt = 0, indix;
for(int i = 0; i < 9; i++) cin>>a[i];
for(int i = 0; i < 8; i++){
bool ok = true;
for(int j = 0; j < 9; j++){
if(a[j] && a[j] != (int)(A[i][j]-'0')) {
ok = false; break;
}
}
if(ok) {cnt++; indix = i;}
}
if(cnt > 1) cout<<"Too Many\n";
else for(int i = 0; i < 10; i++){
cout<<A[indix][i];
if(i % 3 == 2) cout<<endl;
else cout<<" ";
}
}
/*
#define _for(i,a,b) for(i = (a); i <= (b); i++)
int x[9],xx[9];
bool judge(){
memcpy(xx,x,sizeof(x));
sort(xx,xx+9);
for(int i = 0; i < 9; i++) if(xx[i] != i+1) return false;
if(x[6]+x[7] + x[8] != 15 || x[2]+x[5] + x[8] != 15 || x[2]+x[4] + x[6] != 15) return false;
return true;
}
void print(){
for(int i = 0; i < 9; i++)
cout<<x[i];
cout<<endl;
}
int main()
{
_for(x[0],0,9){
_for(x[1],0,9){
_for(x[3],0,9){
_for(x[4],0,9){
x[2] = 15-x[0]-x[1];
x[5] = 15-x[3]-x[4];
x[6] = 15-x[0]-x[3];
x[7] = 15-x[1]-x[4];
x[8] = 15-x[0]-x[4];
if(judge() ) print();
}
}
}
}
return 0;
}
276951438
294753618
438951276
492357816
618753294
672159834
816357492
834159672
*/
excel地址
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
int main(){
int n;
string s;
cin>>n;
while(n>0){
s += char((--n)%26+'A');
n = n/26;
}
reverse(s.begin(),s.end());
cout<<s;
return 0;
}
/*
input
2054
output
BZZ
*/
日期问题
//注意重复
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
int A[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int y, m, d, k = 0;
char s[3][20];
string ss[3];
bool is_rightday(int yy,int mm, int dd){
if(yy >= 60) yy += 1900;
else yy += 2000;
if( (mm < 1) || (mm > 12) ) return false;
if(yy % 4) A[2] = 28;
else A[2] = 29;
if(dd > A[mm] || dd < 1) return false;
sprintf(s[k],"%d-%02d-%02d\n",yy,mm,dd);
return true;
}
int main(){
int a, b, c;
scanf("%d/%d/%d",&a,&b,&c);
if( is_rightday(a,b,c) )k++;
if( is_rightday(c,b,a) )k++;
if( is_rightday(c,a,b) )k++;
for(int i = 0; i < k; i++) ss[i] = s[i];
sort(ss,ss+k);
int n = unique(ss,ss+k) - ss;
for(int i = 0; i < n; i++)
cout<<ss[i];
return 0;
}
最大子阵
//别人的思想 混杂的代码 自己的知识
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn = 500+5;
int M[maxn][maxn], C[maxn][maxn];
int n, m;
int a[maxn], dp[maxn];
int main(){
cin>>n>>m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
scanf("%d",&M[i][j]);
C[i][j] = M[i][j] + C[i-1][j];
}
}
int ans = M[1][1];
for(int i = 0; i < n; i++){
for(int j = i+1; j <= n; j++){ //以下为i+1到j行最大方阵求解
for(int k = 1; k <= m; k++){
dp[k] = C[j][k]-C[i][k];
if(dp[k-1] > 0) dp[k] += dp[k-1];
ans = max(ans, dp[k] );
}
}
}
cout<<ans<<endl;
return 0;
}
/*
input
3 3
-1 -4 3
3 4 -1
-5 -2 8
output
10
*/
油漆问题
//当maxn为8000+50为80分 为10000+50时为0分 待完善
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn = 10000+50;
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
int X1[maxn], Y1[maxn], X2[maxn], Y2[maxn];
int x[maxn*2], y[maxn*2];
int color[maxn][maxn];
bool v[maxn][maxn];
int n, nx, ny;
int ans = 0;
int part(int *x){
sort(x,x+2*n+2);
return unique(x,x+2*n+2)-x;
}
int ID(int a, int n, int *x){
return lower_bound(x, x+n, a) - x;
}
int main(){
cin>>n;
x[0] = y[0] =0, x[1] = y[1] = maxn;
for(int i = 1; i <= n; i++){
scanf("%d%d%d%d",&X1[i],&Y1[i],&X2[i],&Y2[i]);
x[2*i] = ++X1[i]; x[2*i+1] = ++X2[i];
y[2*i] = ++Y1[i]; y[2*i+1] = ++Y2[i];
}
nx = part( x );
ny = part( y );
for(int i = 1; i <= n; i++){
int xx1 = ID(X1[i],nx,x), xx2 = ID(X2[i],nx,x);
int yy1 = ID(Y1[i],ny,y), yy2 = ID(Y2[i],ny,y);
for(int j = xx1; j < xx2; j++)
for(int k = yy1; k < yy2; k++)
color[j][k] = 1;
}
for(int i = 0; i < nx; i++)
for(int j = 0; j < ny; j++)
if(color[i][j])
ans += ( x[i+1]-x[i] )*( y[j+1]-y[j] );
cout<<ans<<endl;
return 0;
}
/*
input
3
1 5 10 10
3 1 20 20
2 7 15 17
---------------------
3
5 2 10 6
2 7 12 10
8 1 15 15
output
340
--------------------
128
*/