版权声明：转载时 别忘了注明出处 https://blog.csdn.net/ZCY19990813/article/details/89069949
只是想记录一下做过的题
比赛总地址 http://acm.sdibt.edu.cn/vjudge/contest
2019省赛训练-1
Sample Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Sample Output
C
C
P
C
C
P
P
C题意：• Suppose m2 + n 2 > 0. Then, 〈m, n〉 is a divisor of 〈p, q〉 if and only if the integer m2 + n 2 is a common divisor of mp + nq and mq − np.就是根据这个公式求出除〈1, 0〉, 〈0, 1〉, 〈−1, 0〉, 〈0, −1〉, 〈m, n〉, 〈−n, m〉, 〈−m, −n〉 and 〈n, −m〉之外有没有其他满足的条件，有就输出C,没有就输出P.
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
#define fori(a,b) for(LL i=a;i<=b;i++)
#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e6+7;
int main()
{
//freopen("//home//acm//桌面//in","r",stdin);
LL t;
cin>> t ;
while(t--){
LL m,n,a,b,c;
cin >> m >> n;
LL max1=max(fabs(m),fabs(n));
LL sum=0;
for(LL i=-max1;i<30000;i++){
for(LL j=-max1;i*i+j*j<20000;j++){
if(i*i+j*j>0){
LL x=i,y=j;
a=x*m+y*n;
b=x*n-y*m;
c=x*x+y*y;
if(a%c==0&&b%c==0&&c!=m*m+n*n&&c!=1){
sum++;
}
if(sum>1)
break;
}
}
if(sum>1)
break;
}
if(sum==0)
cout << "P" << endl;
else
cout << "C" << endl;
}
return 0;
}2019省赛训练-2
Sample Input
ABCA BACA
ELLY KRIS
AAAA ZZZZ
Sample Output
0
29
100题意：给出两个字符串，各自元素可以交换，然后从第一个串到第2个串最少需要几步，A-Z 25步，Z-A 1步。
思路：每个字符串都先按字符串大小排一下序，将第二个串最后一个字符移到前面来，计算一下需要几步，这样经过n(字符串长度)次，求出最小值。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL mod=1e9+7;
int a[111111];
int cmp(char p,char q){
return p<q;
}
int main()
{
char s1[55],s2[55];
while(cin >> s1 >> s2){
int len = strlen(s1);
sort(s1,s1+len,cmp);
sort(s2,s2+len,cmp);
int min1=MAX,sum=0;
for(int i=0;i<len;i++){
sum=0;
for(int j=0;j<len;j++)
{
if(s1[j]>s2[j])
{
sum+=26-(s1[j]-'A'+1)+(s2[j]-'A'+1);
}
else
{
sum+=(s2[j]-'A'+1)-(s1[j]-'A'+1);
}
}
if(sum<=min1)
min1=sum;
char c=s2[0];
for(int j=1;j<len;j++){
s2[j-1]=s2[j];
}
s2[len-1]=c;
}
cout<<min1<<endl;
}
return 0;
}J. Smallest Difference
Input
2
3
1 2 3
5
2 2 3 4 5
OUt
2
3题意：任意挑选x个数，使得这x个数任意两个差的绝对值小于等于1,问x最多有几个。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e5+7;
const LL mod=1e9+7;
int a[M],b[M],c[M];
set <LL> se;
int main()
{
// freopen("//home//acm//桌面//in","r",stdin);
LL t;
cin >> t;
while(t--)
{
int n;
cin >> n;
se.clear();
memset(b,0,sizeof(b));
for(int i=0;i<n;i++){
cin>>a[i];
b[a[i]]++;
se.insert(a[i]);
}
memset(c,0,sizeof(c));
set<LL>::iterator it;
LL e=0;
for(it=se.begin();it!=se.end();it++)
c[e++]=*it;
LL maxx=b[c[0]];
for(int i=1;i<e;i++)
{
LL sum=0;
if(c[i]-c[i-1]<=1)
sum=b[c[i]]+b[c[i-1]];
if(sum>=maxx)
maxx=sum;
if(b[c[i]]>=maxx)//第一次没有想到（样例 0 1 3 3 3）
maxx=b[c[i]];
}
cout<<maxx<<endl;
}
return 0;
}
2019省赛训练-4
Lucky 7
Sample Input
4
3 7
4 5 6
3 7
4 7 6
5 2
2 5 2 5 2
4 26
100 1 2 4
Sample Output
No
Yes
Yes
Yes
题意：给出两个数n,b,问接下来的n个数有没有存在加上b能整除7的。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e5+7;
const LL mod=1e9+7;
LL a[M];
map<LL,LL>m;
int main()
{
//freopen("//home//acm//桌面//in","r",stdin);
int t,n,x,b;
cin>>t;
while(t--)
{
int flag=0;
cin>>n>>b;
for(int i=0;i<n;i++){
cin>>x;
if((x+b)%7==0){
flag=1;
}
}
if(flag)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
}B - King of Karaoke
Sample Input
2
4
1 2 3 4
2 3 4 6
5
-5 -4 -3 -2 -1
5 4 3 2 1
Sample Output
3
1题意：有两组数，问加上一个x，使得两组数里面相同的最多，求这个相同最多的个数。
思路：直接计算对应两个的差，放到map中，在求个数的最大值。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e5+7;
const LL mod=1e9+7;
int a[M],b[M];
map<int,int>m;
int main()
{
//freopen("//home//acm//桌面//in","r",stdin);
int t,n;
cin>>t;
while(t--)
{
cin>>n;
m.clear();
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i++){
cin>>b[i];
int x=b[i]-a[i];
m[x]++;
}
map<int,int>::iterator it;
int maxx=-1;
for(it=m.begin();it!=m.end();it++)
{
if(it->second>=maxx)
maxx=it->second;
}
cout<<maxx<<endl;
}
return 0;
}A - Peak
Sample Input
7
5
1 5 7 3 2
5
1 2 1 2 1
4
1 2 3 4
4
4 3 2 1
3
1 2 1
3
2 1 2
5
1 2 3 1 2
Sample Output
Yes
No
No
No
Yes
No
No题意：问有没有存在这样的情况，前面是单调递增的，后面是单调递减的。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e5+7;
const LL mod=1e9+7;
LL a[M],b[M];
map<LL,LL>m;
int main()
{
//freopen("//home//acm//桌面//in","r",stdin);
LL t;
scanf("%d",&t);
while(t--){
LL n,in=0,de=0,flag=0;
scanf("%d",&n);
for(LL i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(LL i=1;i<=n;i++){
if(i>1&&a[i]<a[i-1]){
flag=1;
}
if(a[i]>=a[i-1]&&i>1)
{
if(a[i]>a[i-1])
in++;
if(flag){
flag=0;
break;
}
}
}
if(flag&&in)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}J - CONTINUE...?
Sample Input
5
1
1
2
10
3
101
4
0000
7
1101001
Sample Output
-1
-1
314
1221
3413214题意：0代表女生，1代表男生，分成4个组，G1,G2,G3,G4 ，要求G1,G2 放女生，其他两个放男生，每个的价值是在字符串的位置，现在要求是否可以分成组，使得G1 G3价值的和等于其他两个的和。
思路：和为奇数不可能分组，偶数时分成和的一半.
(卡了好多次…………)
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
#define MAX 0x3f3f3f3f
//#define fori(a,b) for(LL i=a;i<=b;i++)
//#define forj(a,b) for(LL j=a;j<=b;j++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const LL M=1e5+7;
const LL mod=1e9+7;
char a[M];
LL b[M];
int main()
{
LL t;
scanf("%lld",&t);
while(t--)
{
mem(b,0);
LL n;
LL sum=0;
scanf("%lld",&n);
// getchar();
scanf("%s",a);
sum=(n+1)*n/2;
if(sum%2==1){
printf("-1\n");
}
else {
sum/=2;
for(LL i=n-1;i>=0;i--){
if(sum==0)
break;
if(sum>=i+1){
b[i]=1;
sum-=i+1;
}
else {
b[sum-1]=1;
sum=0;
break;
}
}
for(LL i=0;i<n;i++){
if(b[i]){
if(a[i]=='0'){
printf("1");
}
else
printf("3");
}
else {
if(a[i]=='0')
printf("2");
else
printf("4");
}
}
printf("\n");
}
}
return 0;
}