はい。
AtCoder Beginner Contest 059
風邪かも。
A - Three-letter acronym
#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;
string a,b,c;
cin >> a >> b >> c;
printf("%c",a.at(0)-'a'+'A');
printf("%c",b.at(0)-'a'+'A');
printf("%c",c.at(0)-'a'+'A');
pp;
return 0;
}
変換方法がよくわからないので結構無理やりに。
B - Comparison
#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;
string a,b;
cin >> a >> b;
if(a==b){
cout << "EQUAL" << endl;
}else{
if(a.size()>b.size()) cout << "GREATER" <<endl;
else if (a.size()<b.size()) cout << "LESS" <<endl;
else {
rep(i,a.size()){
if((a.at(i)-'0')>(b.at(0)-'0')) {cout << "GREATER" << endl; break;}
else if((a.at(i)-'0')<(b.at(0)-'0')) {cout << "LESS" << endl; break;}
}
}
}
return 0;
}
文字列扱いで受け取りました。長さ違えばそれだけで大小決定。長さ同じなら先頭、左からみると数の大小判別に都合がよく。
C - Sequence
#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;
ll n,k,x=0,y=0,z=0,cnt=0ll,ans=0ll;
cin >> n;
vector<int> a(n);
rep(i,n) cin >> a.at(i);
rep(i,n){
x+=a.at(i);
if(i%2==0 && x<=0ll) {cnt+=(abs(x)+1); x=1ll;}
else if(i%2==1 && x>=0ll) {cnt+=(x+1); x=-1ll;}
}
x=0;
rep(i,n){
x+=a.at(i);
if(i%2==1 && x<=0) {ans+=(abs(x)+1); x=1ll;}
else if(i%2==0 && x>=0) {ans+=(x+1); x=-1ll;}
}
cout << ((ans<=cnt)?ans:cnt) << endl;
return 0;
}
正負正負正負...か負正負正負...のパターンで作るのを試して比較