OJ系统答案

1.

#include using namespace std; int main() {int n,m; cin>>n;

for (;n>0;n--) {cin>>m;

int t,s=0,p,q=0; for (t=1;t<=m;) {s=s+1; t=t+1;

if (t%7==0) {s=s+1; q=q+1;}} p=s-7*q;

cout<

2,

#include using namespace std; int main() {int n; cin>>n;

for (;n>0;n--)

{int year,month,days;

int sum=0;

cin>>year>>month>>days; for (int m=1;month>m;m++) {switch(m)

{case 1:case 3:case 5:case 7:case 8:case 10: sum=sum+31;break;

case 4:case 6:case 9:case 11: sum=sum+30;break;

case 2:if ((year%4==0)&&(year0!=0)||(year@0==0)) sum=sum+29; else sum=sum+28;}} sum=sum+days;

cout<

3,

#include using namespace std; int main() {char n;

int a=0,b=0,c=0,d=0; cin.get(n);

for (;n!='\\n';)

{if (n>='A'&&n<='Z'||n>='a'&&n<='z') a++;

else if (n==32) b++;

else if (n>='0'&&n<='9') c++; else d++; cin.get(n);}

cout<

4.

#include using namespace std; int A(int); int main() {int n; cin>>n;

for (;n>0;n--) {int a,b,c=1,m=0; cin>>a;

for (;a>0;a--) {cin>>b; m=m+b;

c=c*A(b);} c=A(m)/c;

cout<

for (;a>0;a--) {m=m*a;} return m;}

5.

#include #include

double D(double,double,double,double); using namespace std; int main() {int T; cin>>T;

double XP,YP,XA,YA,XB,YB; for (;T>0;T--)

{cin>>XP>>YP>>XA>>YA>>XB>>YB; double p,s,a,b,c,d; a=D(XP,YP,XA,YA); b=D(XP,YP,XB,YB); c=D(XA,YA,XB,YB);

if (a>b&&(b*b+c*c)

else if (b>a&&(a*a+c*c)

p=(a+b+c)/2;

s=sqrt(p*(p-a)*(p-b)*(p-c)); d=2*s/c;}

cout<

double D(double a,double b,double c,double d) {return sqrt((a-c)*(a-c)+(b-d)*(b-d));}