Codeforces Round #226 (Div. 2) 解题报告

Problem A Bear and Raspberry

题意：水题。找前一天减后一天的差值最大。然后大于c就减去c，否则为0.

代码如下：

//2014-01-24-23.22
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;

int main()
{
//    freopen("in.txt", "r", stdin);

    int n, m, a[110];
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        for(int i=0; i<n; i++){
            scanf("%d", &a[i]);
        }
        int Max = 0;
        for(int i=0; i<n-1; i++){
            Max = max(Max, a[i]-a[i+1]);
        }
        if(Max-m>0)printf("%d
", Max-m);
        else printf("0
");
    }
    return 0;
}

Problem B Bear and Strings

题意：让你找一个字符串中有多少个包含bear的子串。

思路：先递推dp[i]表示到i结束的字符串从i前面几个开始包含bear。然后累加求和就行了。

代码如下：

//2014-01-24-23.22
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 100000+10;

int main()
{
//    freopen("in.txt", "r", stdin);

    int len, tag, dp[LEN];
    char str[LEN];
    while(scanf("%s", str)!=EOF){
        len = strlen(str);
        tag = -1;
        dp[0] = dp[1] = dp[2] = -1;
        for(int i=3; i<len; i++){
            if(str[i]=='r' && str[i-1]=='a' && str[i-2]=='e' && str[i-3]=='b'){
                tag = 3;
            }else if(tag>=3) tag++;
            dp[i] = tag;
        }
        ll ans = 0;
        for(int i=0; i<len; i++){
            if(dp[i]!=-1)ans+=(i-dp[i]+1);
        }
        printf("%I64d
", ans);
    }
    return 0;
}

Problem C Bear and Prime Numbers

题意：有一个数组，定义了一个值f（i）i为素数=数组中有多少个元素能除进i。然后问你一个区间段的f值之和。

思路：先求出所有的f值然后查询。f值得求法是先打一张1E7的素数表，然后每个素数去试除，就ok了。

代码如下：

//2014-01-24-23.22
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAX = 10000000+10;
const int LEN = 1000000+10;
bool isprime[MAX];
int tot;
int a[LEN], num[MAX], prime[MAX];

void makePrime()
{
//    memset(prime, 0, sizeof prime);
    tot = 0;
    for(int i=2; i<=MAX; i++){
        if(!isprime[i]) prime[tot++] = i;
        for(int j=0, k; j<tot && (k=i*prime[j])<=MAX; j++){
            isprime[k] = true;
            if(i%prime[j]==0)break;
        }
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);
    makePrime();
    int n, q;
    while(scanf("%d", &n)!=EOF){
        for(int i=0; i<n; i++) scanf("%d", &a[i]);
        for(int i=0; i<n; i++){
            for(int j=0; j<tot; j++){
                if(prime[j]*prime[j]>a[i])break;
                if(a[i]%prime[j]==0){
                    num[prime[j]]++;
                    while(a[i]%prime[j]==0)a[i]/=prime[j];
                }
            }
            num[a[i]] += (isprime[a[i]]==true?0:1);
        }
        for(int i=1; i<MAX; i++) num[i] += num[i-1];
        int q, l, r;
        scanf("%d", &q);
        for(int i=0; i<q; i++){
            scanf("%d%d", &l, &r);
            l = min(l, MAX-1);
            r = min(r, MAX-1);
            printf("%d
", num[r]-num[l-1]);
        }
    }
    return 0;
}