浙江省赛2019

F - Abbreviation

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int eps=1e-9;
const ll mod=1e9+7;
int rd() {
  int x=0,w=1;
  char ch=0;
  while (ch<'0'||ch>'9'){if (ch == '-') w=-1;ch = getchar();}
  while (ch>='0'&& ch <='9'){x=x * 10+(ch-'0');ch = getchar();}
  return x*w;
}
const int maxn=1e6;
int main(){
    int t;
    
    scanf("%d",&t);
    while(t--){
    string ans,s;
    cin>>s;
    int len=s.size();
    ans+=s[0];
    for(int i=1;i<len;i++){
    if(s[i]=='a'||s[i]=='e'||s[i]=='i'||
    s[i]=='y'||s[i]=='o'||s[i]=='u') 
    continue;
    ans+=s[i];
    
    }
    
    cout<<ans<<endl;
    }
    
    return 0;    
}

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G - Lucky 7 in the Pocket

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int eps=1e-9;
const ll mod=1e9+7;
int rd() {
  int x=0,w=1;
  char ch=0;
  while (ch<'0'||ch>'9'){if (ch == '-') w=-1;ch = getchar();}
  while (ch>='0'&& ch <='9'){x=x * 10+(ch-'0');ch = getchar();}
  return x*w;
}
const int maxn=1e5+5;

int main(){
    int n,t;
    scanf("%d",&t);
    while(t--){
    scanf("%d",&n);
    int m=n;
    for(m=n; ;m++){
    if(m%7==0&&m%4!=0){
    break; 
    }
    
    }
    printf("%d\n",m);
    }
    
    return 0;    
}

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I - Fibonacci in the Pocket

题意：就是给你a，b；然后让你求二者之间斐波那契之和的奇偶；但数据达到了1e1000

考虑这样做，斐波那契是奇数，，奇数，偶数；然后只要看看a，b在第几位就好了；

1 1 或者2 2或者2 0 或者0 1 就是1；其他情况都是0；

这里我犯了一个傻逼错误，就是3%3肯定是0呀；注意一下就好了

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int eps=1e-9;
const ll mod=1e9+7;
int rd() {
  int x=0,w=1;
  char ch=0;
  while (ch<'0'||ch>'9'){if (ch == '-') w=-1;ch = getchar();}
  while (ch>='0'&& ch <='9'){x=x * 10+(ch-'0');ch = getchar();}
  return x*w;
}
const int maxn=1e6;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
    string a,b;
    cin>>a>>b;
    ll sum1=0,sum2=0,x,y;
    int len1=a.size();
    for(int i=0;i<len1;i++){
    sum1+=a[i]-'0';
    }

    int len2=b.size();
    for(int i=0;i<len2;i++){
    sum2+=b[i]-'0';
    }
    if(sum1>sum2)swap(sum1,sum2);
    x=sum1%3;
    y=sum2%3;
//    cout<<"test:"<<endl; 
//    cout<<sum1<<" "<<sum2<<endl;
        if(x==1&&y==1&&sum2>sum1
    ||x==2&&y==2&&sum2>sum1||x==2&&y==0
    ||x==0&&y==1)puts("1");
    else puts("0");
 
    } 
    
    return 0;    
}

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待更新；

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