はい。
AtCoder Beginner Contest 045
喉痛いかも?またですか。。。
A - 台形
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
#define per(i,n) for(int i=n-1;i>=0;--i)
#define sc1(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define sc3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sl1(a) scanf("%lld",&a)
#define sl2(a,b) scanf("%lld %lld",&a,&b)
#define sl3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define PI 3.1415926535897932
#define print(a) cout << a << endl
#define pp puts("")
#define Yes printf("Yes\n")
#define No printf("No\n")
void yneso(int a) {if(a) Yes; else No;}
typedef long long ll;
int souwa(int a) {return (1+a)*a/2;}
int lcm(int a,int b) { return a*b/__gcd(a,b); }
double tilt(int x1,int y1,int x2,int y2) {return (1.0*y2-1.0*y1)/(1.0*x2-1.0*x1);}
double tri(int xa,int ya,int xb,int yb,int xc,int yc) {return (1.0*xa-1.0*xc)*(1.0*yb-1.0*yc)-(1.0*xb-1.0*xc)*(1.0*ya-1.0*yc);}
bool sankaku(int a,int b,int c) {vector <int> t={a,b,c};sort(t.begin(),t.end()); return t.at(0)+t.at(1)>t.at(2);};
int main(){
int a,b,h;
cin >> a >> b >> h;
print((a+b)*h/2);
return 0;
}
台形の面積を問題文の式の通りに。
B - 3人でカードゲームイージー
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
#define per(i,n) for(int i=n-1;i>=0;--i)
#define sc1(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define sc3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sl1(a) scanf("%lld",&a)
#define sl2(a,b) scanf("%lld %lld",&a,&b)
#define sl3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define PI 3.1415926535897932
#define print(a) cout << a << endl
#define pp puts("")
#define Yes printf("Yes\n")
#define No printf("No\n")
void yneso(int a) {if(a) Yes; else No;}
typedef long long ll;
int souwa(int a) {return (1+a)*a/2;}
int lcm(int a,int b) { return a*b/__gcd(a,b); }
double tilt(int x1,int y1,int x2,int y2) {return (1.0*y2-1.0*y1)/(1.0*x2-1.0*x1);}
double tri(int xa,int ya,int xb,int yb,int xc,int yc) {return (1.0*xa-1.0*xc)*(1.0*yb-1.0*yc)-(1.0*xb-1.0*xc)*(1.0*ya-1.0*yc);}
bool sankaku(int a,int b,int c) {vector <int> t={a,b,c};sort(t.begin(),t.end()); return t.at(0)+t.at(1)>t.at(2);};
int main(){
int mod=1e9+7;
int x=0,y=0,z,cnt=0,ans=0;
string a,b,c;
cin >> a >> b >>c;
for(;;){
if(x==0 && a.size()==0) {print("A"); break;}
if(x==1 && b.size()==0) {print("B"); break;}
if(x==2 && c.size()==0) {print("C"); break;}
if(x==0) {y=a.at(0)-'a'; a=a.substr(1);}
else if(x==1) {y=b.at(0)-'a'; b=b.substr(1);}
else {y=c.at(0)-'a'; c=c.substr(1);}
x=y;
}
return 0;
}
substrで文字列操作。空で手番が来たら勝利。
C - たくさんの数式
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
#define per(i,n) for(int i=n-1;i>=0;--i)
#define sc1(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define sc3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sl1(a) scanf("%lld",&a)
#define sl2(a,b) scanf("%lld %lld",&a,&b)
#define sl3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define PI 3.1415926535897932
#define print(a) cout << a << endl
#define pp puts("")
#define Yes printf("Yes\n")
#define No printf("No\n")
void yneso(int a) {if(a) Yes; else No;}
typedef long long ll;
int souwa(int a) {return (1+a)*a/2;}
int lcm(int a,int b) { return a*b/__gcd(a,b); }
double tilt(int x1,int y1,int x2,int y2) {return (1.0*y2-1.0*y1)/(1.0*x2-1.0*x1);}
double tri(int xa,int ya,int xb,int yb,int xc,int yc) {return (1.0*xa-1.0*xc)*(1.0*yb-1.0*yc)-(1.0*xb-1.0*xc)*(1.0*ya-1.0*yc);}
bool sankaku(int a,int b,int c) {vector <int> t={a,b,c};sort(t.begin(),t.end()); return t.at(0)+t.at(1)>t.at(2);};
int main(){
int x,y,z;
ll cnt=0ll,ans=0ll;
string s;
cin >> s;
for(int i=0;i<(1<<s.size()-1);i++) {
cnt=s.at(0)-'0';
x=i;
for(int j=1;j<s.size();j++){
if (x&1) {ans+=cnt; cnt=0;}
else {cnt*=10;}
cnt+=s.at(j)-'0';
x=x>>1;
}
ans+=cnt;
}
print(ans);
return 0;
}
bitで+使う使わないをよしなに全列挙できればよしなに。