Created
July 24, 2016 21:44
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/** | |
La idea en este problema es llevar un acumulado pero de los numeros que hay (del 0 al 9) hasta x posición | |
**/ | |
#include <iostream> | |
#include <cmath> | |
#include <algorithm> | |
#include <queue> | |
#include <stack> | |
#include <sstream> | |
#include <map> | |
#include <cstdio> | |
#include <queue> | |
#include <string> | |
#include <iomanip> | |
#define gc getchar_unlocked | |
using namespace std; | |
typedef vector<int> vi; | |
int n,q,l,r; | |
int N[100005][10]; | |
/**solo es para leer más rapido, se puede usar scanf en su lugar**/ | |
void scanint(int &x) | |
{ | |
register int c = gc(); | |
x = 0; | |
int neg = 0; | |
for(;((c<48 || c>57) && c != '-');c = gc()); | |
if(c=='-') {neg=1;c=gc();} | |
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;} | |
if(neg) x=-x; | |
} | |
int aux; | |
int main(){ | |
while(scanf("%d",&n)!=EOF){ | |
for(int i=1;i<=n;i++){ | |
scanint(aux); | |
for(int e=0;e<10;e++) | |
N[i][e]=N[i-1][e]; | |
N[i][aux]++; | |
} | |
scanint(q); | |
while(q--){ | |
scanf("%d %d",&l,&r); | |
if(l==r){ | |
printf("1\n"); | |
continue; | |
} | |
l--; | |
int res=0; | |
for(int i=0;i<10;i++){ | |
res+=((N[r][i]-N[l][i])>=1?1:0); | |
} | |
printf("%d\n",res); | |
} | |
} | |
return 0; | |
} |
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