Codeforces Round #308 (Div. 2)

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A - Vanya and Table 每次都给出一个子矩阵并让+1，最后求矩阵内数之和

#include <bits/stdc++.h>using namespace std;int main(){    int n, x1, y1, x2, y2, s = 0;    scanf("%d",&n);    while( n-- )    {        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);        s = s + (x2 - x1 + 1) * (y2 - y1 + 1);    }    printf("%d\n", s);    return 0;}

B - Vanya and Books1~n的数里面有多少位。暴力模拟

#include <bits/stdc++.h>using namespace std;typedef long long lol;int main() {    lol n;    while ( cin >> n ) {        lol temp = 1000000000;        lol cnt = 10;        lol res = 0;        while ( temp>0&&n>0 ) {            if ( n-(temp-1)>0 ) {                res += (n-(temp-1))*(cnt);                n = temp - 1;            }            temp /= 10;            -- cnt;        }        cout << res << endl;    }    return 0;}

C - Vanya and Scales给出w进制的秤和m重的东西，问能否用秤称出m重的东西。

m化成w进制的数，则表示成 p0 * w^0 + p1 * w^1 + p2 * w^2.......px * w^x。取其中一项pc * w^c，若pc = w-1，则pc * w^c + w^c = w * w^（c+1），即若最开始m重的东西放在左边，w进制里面若有一项系数为w-1，即等价于在左边放一个 w^c重的秤砣，右边放一个 w^(c+1)的秤砣。其他情况下，pc系数为0或1表示右边放或不放，是满足的

int main(){    int dig[222], tot;    ll m, w;    while( ~scanf("%lld%lld", &w, &m) ) {        memset( dig, 0, sizeof( dig ) );        tot = 0;        ll x = m;        while( x ) {            dig[tot++] = x % w;            x /= w;        }        bool OK = 1;        for( int i = 0; i < tot; ++i ) {            if( dig[i] == 0 || dig[i] == 1 )                continue;            if( dig[i] >= w-1 ) {                dig[i+1]++;                tot++;            }            else {                OK = 0;                break;            }        }        puts( OK? "YES" : "NO" );    }    return 0;}

D - Vanya and Triangles给出平面上n个点，问能组成多少个三角形。

枚举每个点出发别的点到他的斜率即可

const int N = 2020;ll x[N], y[N];ll n;map <ll, ll> mp;map <ll, ll> :: iterator it;pair <ll, ll> f( int i, int j ){    pair <ll, ll> re;    ll xx = x[i] - x[j], yy = y[i] - y[j];    if( xx == 0 ) {        re.first = 1LL*N, re.second = 1LL*N;        return re;    }    if( yy == 0 ) {        re.first = 1LL*0, re.second = 1LL*0;        return re;    }    ll xxx = xx / abs(xx), yyy = yy / abs(yy);    yyy *= xxx, xxx = 1;    xx = abs(xx), yy = abs(yy);    ll gcd = __gcd( xx, yy );    re.first = xx / gcd * xxx, re.second = yy / gcd * yyy;    return re;}int main(){    while( ~scanf("%lld", &n) ) {        mp.clear();        ll ans = n * (n-1) * (n-2) / 6, tmp = 0;        for( int i = 1; i <= n; ++i )            scanf("%lld %lld", &x[i], &y[i]);        for( int i = 1; i <= n; ++i ) {            mp.clear();            for( int j = 1; j <= n; ++j ) {                if( i == j )                    continue;                pair <ll, ll> c = f( i, j );                ll xx = c.first, yy = c.second;                mp[xx*N+yy]++;            }            for( it = mp.begin(); it != mp.end(); ++it ) {                tmp += (it->second >= 2) ? ((it->second-1) * it->second / 2) :0;            }        }        ans -= tmp / 3;        printf("%lld\n", ans);    }    return 0;}

Vanya and Brackets

表达式加一对括号，求表达式的最大值

括号加在*旁边就行了，暴力模拟过去，由于*最多15，复杂度15*15*n

#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <string>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <iomanip>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8#define asd puts("sdfsdfsdfsdfsdfsdf");typedef long long ll;///typedef __int64 LL;const int inf = 0x3f3f3f3f;const int N = 5050;int p[20], pos;char s[N];int main(){    while( ~scanf("%s", s+1 ) ) {        int len = strlen( s+1 );        s[len+1] = '+';        pos = 0, p[pos++] = 0;        for( int i = 1; i <= len+1; ++i ) {            if( s[i] == '*' )                p[++pos] = i;        }        p[++pos] = len+1;        ll ans = 0;        if( pos == 2 ) {            ans = s[1] - '0';            for( int i = 2; i <= len+1; ++i ) {                if( s[i] != '+' )                    ans += s[i] - '0';            }            printf("%lld\n", ans);            continue;        }        for( int i = 0; i < pos; ++i ) {            for( int j = i; j <= pos; ++j ) {                int l = p[i], r = p[j];                ll x = 0, tmp = 1, op = 0;      //op = 0----+  op = 1 ----- *                for( int k = 1; k <= l; ++k ) {                    if( s[k] >= '0' && s[k] <= '9' ) {                        if( op == 0 )                            tmp = s[k] - '0';                        else                            tmp *= s[k] - '0';                    }                    else if( s[k] == '+' ) {                        x += tmp;                        op = 0;                    }                    else {                        op = 1;                    }                }                //printf("%d %d %lld %lld\n", l, r, x, tmp);                ll y = 0, op1 = 0, tmp1 = 1;                for( int k = l+1; k <= r; ++k ) {                    if( s[k] >= '0' && s[k] <= '9' ) {                        if( op1 == 0 )                            tmp1 = s[k] - '0';                        else                            tmp1 *= s[k] - '0';                    }                    else if( s[k] == '+' ) {                        y += tmp1;                        op1 = 0;                    }                    else {                        op1 = 1;                    }                }                if( op1 == 1 )                    y += tmp1;                //printf("%d %d %lld %lld\n", l, r, y, tmp1);                ll tmp2 = 1;                int o = r+1;                for( ; o <= len+1 && s[o] != '+'; ++o ) {                    if( s[o] >= '0' && s[o] <= '9' )                        tmp2 *= s[o]-'0';                }                //printf("%d %d %lld \n", l, r, tmp2);                x += tmp * y * tmp2;                op = 0, tmp = 0;                for( int k = o+1; k <= len+1; ++k ) {                    if( s[k] >= '0' && s[k] <= '9' ) {                        if( op == 0 )                            tmp = s[k] - '0';                        else                            tmp *= s[k] - '0';                    }                    else if( s[k] == '+' ) {                        x += tmp;                        op = 0;                    }                    else {                        op = 1;                    }                }                //printf("%lld\n\n", x);                ans = max( ans, x );            }        }        printf("%lld\n", ans);    }    return 0;}// 1+2*3+4

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