A 15 and 20
Simulation direct up
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int a1,b1,c1;
scanf("%d%d%d",&a1,&b1,&c1);
int a2,b2,c2;
scanf("%d%d%d",&a2,&b2,&c2);
int sum=b1+c1+b2+c2;
if(sum==a1) {
if(sum==a2)
puts("continue");
else
puts("alice");
} else {
if(sum==a2)
puts("bob");
else
puts("continue");
}
}
return 0;
}B display
Direct simulation, pay attention to details
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int c,r,n;
scanf("%d%d%d",&c,&r,&n);
while(n--) {
int a,b;
scanf("%d%d",&a,&b);
int c1=c/a,c2=r/b;
int sum1=c1*a,sum2=c1*b;
int sum3=c2*a,sum4=c2*b;
if(sum1<=c&&sum2<=r) {
if(sum3<=c&&sum4<=r) {
if(sum1*sum2>=sum3*sum4)
printf("%d %d\n",sum1,sum2);
else
printf("%d %d\n",sum3,sum4);
} else
printf("%d %d\n",sum1,sum2);
} else {
if(sum3<=c&&sum4<=r)
printf("%d %d\n",sum3,sum4);
else
printf("%d %d\n",0,0);
}
}
if(t!=0)
printf("\n");
}
return 0;
}C Sequence
The recurrence formula is
Take three as cycle section
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
long long a1,a2,k;
scanf("%lld%lld%lld",&a1,&a2,&k);
long long a3=a2-a1;
if(k<=3) {
if(k==1) {
printf("%lld\n",a1);
}
if(k==2) {
printf("%lld\n",a2);
}
if(k==3) {
printf("%lld\n",a3);
}
} else {
long long z2=k%3;
if(z2==1) {
long long z1=(k-1)/3;
if(z1%2==0)
printf("%lld\n",a1);
else
printf("%lld\n",-a1);
}
if(z2==2) {
long long z1=(k-2)/3;
if(z1%2==0)
printf("%lld\n",a2);
else
printf("%lld\n",-a2);
}
if(z2==0) {
long long z1=(k-3)/3;
if(z1%2==0)
printf("%lld\n",a3);
else
printf("%lld\n",-a3);
}
}
}
return 0;
}D Carrot
Judge the case of n==2, and then judge the parity of n
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
long long n;
scanf("%lld",&n);
if(n==2)
puts("2");
else {
if(n%2==0)
puts("0");
else
puts("1");
}
}
return 0;
}E Cuboid
Enumeration of violence
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
long long sum1=(b+c)*(b+c)+a*a;
long long sum2=(a+c)*(a+c)+b*b;
long long sum3=(a+b)*(a+b)+c*c;
printf("%lld\n",min(sum1,min(sum2,sum3)));
}
return 0;
}F Order
In consideration of greed, when m pieces are produced every day until they are less than m at last, the answer can be obtained by adding simulation results. It's too late for the first two posts. It's a pity
#include<bits/stdc++.h>
using namespace std;
vector<int> p;
int main() {
int t;
scanf("%d",&t);
while(t--) {
p.clear();
int n,m;
scanf("%d%d",&n,&m);
p.resize(n);
for(int i=0;i<n;i++)
scanf("%d",&p[i]);
int start=1,sum=0;
for(int i=0;i<n;){
int temp=m;
while(i<n&&temp>=p[i]) {
temp-=p[i];
i++;
sum+=start;
}
start++;
}
printf("%d\n",sum);
}
return 0;
}G Ball
The game didn't write out. After the game, the question was found to be the conclusion. There was no one left. It was a pity that I kept pushing for the last half hour
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
long long n,m;
scanf("%lld%lld",&n,&m);
long long d=__gcd(n,m);
n/=d,m/=d;
if(abs(n-m)%2==1)
puts("-1");
else
printf("%lld\n",n+m-2);
}
return 0;
}H strange palindrome substring
It will not be repaired.
I Line
If the rectangle is divided into two parts, then this point must go back to the midpoint on x and y of the rectangle. This point was unexpected at the time of the game, but it's a pity. So we know P point and another point, which we temporarily call the midpoint. Two points determine a straight line, judge whether there is a straight line and simplify it.
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
long long x1,y1,x2,y2,x3,y3;
cin>>x1>>y1>>x2>>y2>>x3>>y3;
long long x4=(x1+x2),y4=y1+y2;
x3*=2,y3*=2;
long long a=y3-y4;
long long b=x4-x3;
long long c=-(a*x3+b*y3);
c/=2;
if(a==0){
long long k=__gcd(b,c);
b/=k,c/=k;
if(b<0)b/=-1,c/=-1;
cout<<a<<' '<<b<<' '<<c<<endl;
continue;
}
long long k=__gcd(a,__gcd(b,c));
a/=k,b/=k,c/=k;
if(a<0)
a/=-1,b/=-1,c/=-1;
printf("%lld %lld %lld\n",a,b,c);
}
return 0;
}J Pythagorean theorem
Thanks to Google and Baidu search engine, we successfully found the construction method of Pythagorean number in the competition, which can be divided into odd number and even number
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
long long a;
scanf("%lld",&a);
if(a%2==1) {
long long z=(a-1)/2;
printf("%lld %lld\n",2*z*z+2*z,2*z*z+2*z+1);
} else {
if(a==4)
printf("3 5\n");
else
printf("%lld %lld\n",(a/2)*(a/2)-1,(a/2)*(a/2)+1);
}
}
return 0;
}K BAD String
It will not be repaired.
