Submission #13473067


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#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;
#define int long long
#define double long double
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x){
	rep(0,i,30){
		if(x<10){
			return i+1;
		}
		x=x/10;
	}
}	
int gcd(int x,int y){
	if(x==0||y==0)return x+y;
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return bb;
		}	
		bb=bb%aa;
		if(bb==0){
			return aa;
		}
	}
}	
int lcm(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return x/bb*y;
		}	
		bb=bb%aa;
		if(bb==0){
			return x/aa*y;
		}
	}
}
bool p(int x){
	if(x==1)return false;
	rep(2,i,sqrt(x)+1){
		if(x%i==0&&x!=i){
			return false;
		}
	}	
	return true;
}
int max(int a,int b){
	if(a>=b)return a;
	else return b;
}
string maxst(string s,string t){
	int n=s.size();
	int m=t.size();
	if(n>m)return s;
	else if(n<m)return t;
	else{
		rep(0,i,n){
			if(s[i]>t[i])return s;
			if(s[i]<t[i])return t;
		}
		return s;
	}
}		
int min(int a,int b){
	if(a>=b)return b;
	else return a;
}	
int n2[41];		
int nis[41]; 
int nia[41];
int mody[41];
int nn;
int com(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int bunsi=1,bunbo=1;
	rep(0,i,y)bunsi=(bunsi*(n-i))%mod;
	rep(0,i,y)bunbo=(bunbo*(i+1))%mod;
	mody[0]=bunbo;
	rep(1,i,41){
		bunbo=(bunbo*bunbo)%mod;
		mody[i]=bunbo;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			bunsi=(bunsi*mody[i])%mod;
		}
	}
	return bunsi;
}
int gyakugen(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	mody[0]=y;
	rep(1,i,41){
		y=(y*y)%mod;
		mody[i]=y;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			n=(n*mody[i])%mod;
		}
	}
	return n;	
}	
int yakuwa(int n){
	int sum=0;
	rep(1,i,sqrt(n+1)){
		if(n%i==0)sum+=i+n/i;
		if(i*i==n)sum-=i;
	}
	return sum;
}	
bool integer(double n){
	if(n==long(n))return true;
	else return false;	
}
int poow(int y,int n){	
	if(n==0)return 1;
	n-=1;
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int yy=y;
	mody[0]=yy;
	rep(1,i,41){
		yy=(yy*yy)%mod;
		mody[i]=yy;
	}
	rep(0,i,41)nis[i]=0;
	nn=n;
	for(int i=40;i>=0;i-=1){
		if(nn>=n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	rep(0,i,41){
		if(nis[i]==1){
			y=(y*mody[i])%mod;
		}
	}
	return y;	
}
int minpow(int x,int y){
	int sum=1;
	rep(0,i,y)sum*=x;
	return sum;
}
int ketawa(int x,int sinsuu){
	int sum=0;
	rep(0,i,100)sum+=(x%poow(sinsuu,i+1))/(poow(sinsuu,i));
	return sum;
}	
int sankaku(int a){
	return a*(a+1)/2;
} 
int same(int a[1111111],int n){
	int ans=0;
	rep(0,i,n){
		if(a[i]==a[i+1]){
			int j=i;
			while(a[j+1]==a[i]&&j<=n-2)j++;
			ans+=sankaku(j-i);
			i=j;
		}
	}
	return ans;
}
bool search(int x){
	int n;
	int a[n];
	int l=0,r=n;
	while(r-l>=1){
		int i=(l+r)/2;
		if(a[i]==x)return true;
		else if(a[i]<x)l=i+1;
		else r=i;
	}
	return false;
}
/*using Graph=vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
    depth[v] = d;
    oya[v]=p;
    for (auto nv : G[v]) {
       if (nv == p) continue; // nv が親 p だったらダメ
       dfs(G, nv, v, d+1); // d を 1 増やして子ノードへ
    }
}*/
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
    memo[h][w] = 'x';

    // 八方向を探索
    for (int dh = -1; dh <= 1; ++dh) {
        for (int dw = -1; dw <= 1; ++dw) {
        	if(abs(0-dh)+abs(0-dw)==2)continue;
            int nh = h + dh, nw = w + dw;

            // 場外アウトしたり、0 だったりはスルー
            if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
            if (memo[nh][nw] == 'x') continue;

            // 再帰的に探索
            dfs(nh, nw);
        }
    }
}*/
int XOR(int a,int b){
	int ni=1;
	rep(0,i,41){
		n2[i]=ni;
		ni*=2;
	}
	rep(0,i,41)nis[i]=0;
	for(int i=40;i>=0;i-=1){
		if(a>=n2[i]){
			nis[i]++;
			a-=n2[i];
		}
		if(b>=n2[i]){
			nis[i]++;
			b-=n2[i];
		}	
	}
	int sum=0;
	rep(0,i,41)sum+=(nis[i]%2*n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2
	
    UnionFind(int N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    
    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
struct aiueo {
	vector<int> par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2
	
    aiueo(int N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    
    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
signed main(){
	sc(s)
	int n=s.size();
	int ans=0;
	rep(0,i,n){
		int j=i+1;
		while(s[j]>s[i]&&j<=n-1){
			j++;
		}
		i=j-1;
		ans++;
	}	
	c(ans)
}

Submission Info

Submission Time
Task D - Concatenation
User tada72
Language C++14 (GCC 5.4.1)
Score 300
Code Size 7216 Byte
Status
Exec Time 5 ms
Memory 512 KB

Test Cases

Set Name Score / Max Score Test Cases
Sample 0 / 0 s1.txt, s2.txt, s3.txt
All 300 / 300 01.txt, 02.txt, 03.txt, 04.txt, 05.txt, 06.txt, 07.txt, 08.txt, 09.txt, 10.txt, 11.txt, 12.txt, 13.txt, 14.txt, 15.txt, 16.txt, 17.txt, 18.txt, 19.txt, 20.txt, 21.txt, 22.txt, 23.txt, 24.txt, 25.txt, 26.txt, 27.txt, s1.txt, s2.txt, s3.txt
Case Name Status Exec Time Memory
01.txt 1 ms 256 KB
02.txt 1 ms 256 KB
03.txt 4 ms 512 KB
04.txt 4 ms 512 KB
05.txt 1 ms 256 KB
06.txt 1 ms 256 KB
07.txt 1 ms 256 KB
08.txt 1 ms 256 KB
09.txt 1 ms 256 KB
10.txt 1 ms 256 KB
11.txt 4 ms 512 KB
12.txt 4 ms 512 KB
13.txt 4 ms 512 KB
14.txt 4 ms 512 KB
15.txt 4 ms 512 KB
16.txt 4 ms 512 KB
17.txt 5 ms 512 KB
18.txt 4 ms 512 KB
19.txt 4 ms 512 KB
20.txt 4 ms 512 KB
21.txt 4 ms 512 KB
22.txt 4 ms 512 KB
23.txt 4 ms 512 KB
24.txt 4 ms 512 KB
25.txt 4 ms 512 KB
26.txt 4 ms 384 KB
27.txt 4 ms 512 KB
s1.txt 1 ms 256 KB
s2.txt 1 ms 256 KB
s3.txt 1 ms 256 KB