BZOJ 4397 [Usaco2015 dec]Breed Counting

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=4397

题意:给你一个只含1,2,3的序列,每次问区间[a,b],输出区间里三个数分别的个数。

思路:开三个数组维护一下前缀和就好了。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define Key_Value ch[ch[root][1]][0]
#define DBN1(a)           cerr<<#a<<"="<<(a)<<"\n"
#define DBN2(a,b)         cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define DBN3(a,b,c)       cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define DBN4(a,b,c,d)     cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define DBN5(a,b,c,d,e)   cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
const int INF=0x3f3f3f3f;
const int P=1000000007;
const double PI=acos(-1.0);
template<typename T>
inline T read(T&x){
    x=0;int _f=0;char ch=getchar();
    while(ch<'0'||ch>'9')_f|=(ch=='-'),ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x=_f?-x:x;
}
inline int add(int a,int b){return (a+=b)>=P?a-P:a;}
inline int sub(int a,int b){return (a-=b)<0?a+P:a;}
inline int mul(int a,int b){return 1LL*a*b%P;}
int n,Q,pre[maxn][3];
int main(){
    read(n),read(Q);
    for (int i=1;i<=n;i++){
        int x;read(x);
        pre[i][0]+=pre[i-1][0]+(x==1);
        pre[i][1]+=pre[i-1][1]+(x==2);
        pre[i][2]+=pre[i-1][2]+(x==3);
    }
    for (;Q--;){
        int a,b;read(a),read(b);
        printf("%d %d %d\n",pre[b][0]-pre[a-1][0],pre[b][1]-pre[a-1][1],pre[b][2]-pre[a-1][2]);
    }
    return 0;
}

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