Codeforces Round #308 (Div. 2) 题解

题目链接：http://codeforces.com/contest/552

A、Vanya and Table

题意：给你n个矩形，问你他们的面积和是多少，是这样的吧-w-

解：~

/*
 * Problem:  
 * Author:  SHJWUDP
 * Created Time:  2015/7/3 星期五 12:19:42
 * File Name: 233.cpp
 * State: 
 * Memo: 
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n;
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%d", &n)) {
        int sum=0;
        for(int i=0; i<n; i++) {
            int a, b, c, d;
            scanf("%d%d%d%d", &a, &b, &c, &d);
            sum+=(c-a+1)*(d-b+1);
        }
        printf("%d
", sum);
    }
    return 0;
}

B、Vanya and Books

题意：给你一个n，问你1~n有多少位数，tips：1,7,5是一位数；21,33,99是两位数

解：~

/*
 * Problem:  
 * Author:  SHJWUDP
 * Created Time:  2015/7/3 星期五 12:19:42
 * File Name: 233.cpp
 * State: 
 * Memo: 
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int64;

const int MaxA=9+7;

int n;
int64 table[MaxA];
int64 pow(int x, int n) {
    int64 res=1;
    while(n--) {
        res*=x;
    }
    return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    table[0]=0;
    for(int i=1; i<=9; i++) {
        table[i]=table[i-1]+(pow(10, i)-pow(10, i-1))*i;
    }
    while(~scanf("%d", &n)) {
        int x=0;
        while(pow(10, x+1)<=n) x++;
        printf("%I64d
", table[x]+(x+1)*(n-pow(10, x)+1));
    }
    return 0;
}

C、Vanya and Scales

题意：给你一个天平和一些砝码，砝码的质量为w的0~100次方（每种一个），问你能不能称出质量为m的物品

解：看到一个很棒的解法，将m化为w进制，从低位到高位考虑，如果该位为0或者1，直接跳过，如果为w-1，将m增加1个当前位的砝码，继续运算，这样就绕过了情况较复杂的减法

/*
 * Problem:  
 * Author:  SHJWUDP
 * Created Time:  2015/7/13 星期一 14:10:32
 * File Name: 233.cpp
 * State: 
 * Memo: 
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MaxA=64+7;

int W, M;
int stk[MaxA], top;
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%d%d", &W, &M)) {
        if(W<=3) {
            puts("YES");
            continue;
        }
        top=0;
        bool ok=1;
        do {
            int x=M%W;
            if(x<=1) M/=W;
            else if(x==W-1) M=M/W+1;
            else {
                ok=0;
                break;
            }
        } while(M);
        puts(ok?"YES":"NO");
    }
    return 0;
}

D、Vanya and Triangles

题意：给你欧式平面上的一些点，问你这些点做顶点能组成多少个三角形

解：~

/*
 * Problem:  
 * Author:  SHJWUDP
 * Created Time:  2015/7/13 星期一 14:41:52
 * File Name: 233.cpp
 * State: 
 * Memo: 
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int MaxA=2000+7;

struct Point {
    int x, y;
} p[MaxA];

int n;
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%d", &n)) {
        for(int i=0; i<n; i++) {
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        int ans=0;
        for(int i=0; i<n; i++) {
            int surplus=n-i-1;
            map<pair<int, int>, int> M;
            for(int j=i+1; j<n; j++) {
                int dx=p[j].x-p[i].x;
                int dy=p[j].y-p[i].y;
                if(dx<0) dx=-dx, dy=-dy;
                int d=__gcd(dx, dy);
                dx/=d; dy/=d;
                pair<int, int> x(dx, dy);
                M[x]++;
            }
            for(const auto& u : M) {
                surplus-=u.second;
                ans+=u.second*surplus;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

E、Vanya and Brackets

题意：给你一个只有+和*还有数字1~9组成的算式，让你加入一对括号使得算式的值尽可能大

解：遇见这么深藏不露的模拟题也是醉了

/*
 * Problem:  
 * Author:  SHJWUDP
 * Created Time:  2015/7/13 星期一 22:17:46
 * File Name: 233.cpp
 * State: 
 * Memo: 
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int64;

const int MaxA=5e3+7;

char str[MaxA];
int64 arr[MaxA], n;
char op[MaxA];  //0:'+' 1:'*'
int64 func(int l, int r) {
    int64 res=0, x=arr[l];
    for(int i=l; i<r; i++) {
        if(op[i]=='+') {
            res+=x;
            x=arr[i+1];
        } else {
            x*=arr[i+1];
        }
    }
    res+=x;
    return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%s", str)) {
        int len=strlen(str);
        n=0;
        vector<int> pos;
        for(int i=0; i+1<len; i+=2) {
            if(i==0 || str[i-1]!=str[i+1]) pos.push_back(n);
            arr[n]=str[i]-'0';
            op[n++]=str[i+1];
        }
        pos.push_back(n);
        arr[n]=str[len-1]-'0';
        int64 ans=func(0, n);
        for(int i=0, lim=(int)pos.size(); i<lim; i++) {
            int64 x=0;
            int64 pm=1;
            int p=pos[i]-1;
            while(p>=0 && op[p]=='*') pm*=arr[p--];
            if(0<=p) x=func(0, p);
            for(int j=i+1; j<lim; j++) {
                int64 y=0;
                p=pos[j];
                int64 am=1;
                while(p<n && op[p]=='*') am*=arr[++p];
                if(p+1<=n) y=func(p+1, n);
                ans=max(ans, x+y+am*pm*func(pos[i], pos[j]));
            }
        }
        printf("%I64d
", ans);
    }
    return 0;
}