NOIP2012 题解

Vigenère 密码

　　这个名字实在打不来...

　　题解：模拟

#include <cstdio>
#include <cstring>

const int MAXL = 1000+10;

bool cj;
int cl, kl;
char c[MAXL], k[MAXL], an;

int main(){
    gets(k), gets(c);
    kl = strlen(k), cl = strlen(c);
    for (int i=0; i<kl; i++) 
        if (k[i]<97) k[i] += 32;

    for (int i=0; i<cl; i++){
        if (c[i]<97) c[i] += 32, cj = true; 
        else cj = false;
        an = (c[i]-'a'-(k[i%kl]-'a')+26)%26+'a';
        if (cj) an -= 32;
        printf("%c", an);
    }
}

 

国王游戏：

　　题解：贪心+高精度，好吧，我看了wikioi的c++最快，这个方法简直....

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::sort;

const int MAXN = 1000+10;
const int MAXL = 100000+10;
const int MOD = 10000;

int n;
double sum[MAXN];

struct bigInteger{
    int len, a[MAXL];

    bigInteger(){
        memset(a, 0, sizeof(a));
        len = 0;
    }

    inline void print(){
        printf("%d", a[len]);
        for (int i=len-1; i>=1; i--)
            printf("%04d", a[i]);
        printf("
");
    }
}Pri;

struct Minister{
    int l, r;
    double sm;

    friend bool operator < (const Minister& a, const Minister& b){
        return a.sm<b.sm;    
    }
}mt[MAXN];

void operator *= (bigInteger& a, int b){
    for (int i=1, c=0, t; i<=a.len+2; i++)
        t = a.a[i]*b+c,
        a.a[i] = t%MOD,
        c = t/MOD;
    
    while (a.a[a.len+1]) a.len ++;
}

void operator /= (bigInteger& a, int b){
    for (int i=a.len, c=0; i>=1; i--)
        c = c*MOD+a.a[i],
        a.a[i] = c/b,
        c %= b;
    while (!a.a[a.len] && a.len>1) -- a.len;
}

int main(){
    scanf("%d", &n);
    for (int i=0; i<=n; i++)
        scanf("%d %d", &mt[i].l, &mt[i].r),
        mt[i].sm = log(mt[i].l)+log(mt[i].r);
     
    sort(mt+1, mt+n+1);

    double t;
    sum[0] = log(mt[0].l);
    for (int i=1; i<=n; i++)
        t = log(mt[i].l),
        sum[i] = sum[i-1]+t;

    int k = 0;
    t = -1;
    for (int i=1; i<=n; i++)
        if (sum[i-1]-log(mt[i].r)>t)
            k = i, t = sum[i-1]-log(mt[i].r);

    Pri.len = 1, Pri.a[1] = 1;
    for (int i=0; i<k; i++)
        Pri *= mt[i].l;
    Pri /= mt[k].r;
    Pri.print();
}

同余方程

　　题解：我只会用扩展欧几里得解...

#include <cstdio>

int a, b, x, y, t;

inline void ExGcd(int a, int b){
    if (b==0) return;
    ExGcd(b, a%b);
    t = x, x = y, y = t-(a/b)*x;
}

int main(){
    scanf("%d %d", &a, &b), x = 1, y = 0;
    ExGcd(a, b);
    while (x<=0) x += b;
    printf("%d", x);
}

借教室

　　题解：二分。开始没加读入优化，在vijos上TLE了...

#include <cstdio>
#include <cstring>

const int MAXN = 1e6+10;

struct Indent{
    int l, r, n;    
}d[MAXN];

int n, m, mid, temp[MAXN], lim[MAXN];
char c;

inline int NextInt(){
    int ret = 0;
    do
        c = getchar();
    while (!(48<=c && c<=57));
    
    do
        ret *= 10, ret += c-48, c = getchar();
    while (48<=c && c<=57);

    return ret;
}

inline bool Check(int x){
    memset(temp, 0, sizeof(temp));

    int sum = 0;
    for (register int i=1; i<=x; i++)
        temp[d[i].l] += d[i].n, temp[d[i].r+1] -= d[i].n;
    for (register int i=1; i<=n; i++){
        sum += temp[i];
        if (sum>lim[i]) return false;
    }
    return true;
}

int main(){
    n = NextInt(), m = NextInt();
    for (register int i=1; i<=n; i++)
        lim[i] = NextInt();
    for (register int i=1; i<=m; i++)
        d[i].n = NextInt(), d[i].l = NextInt(), d[i].r = NextInt();
            
    int l = 1, r = m+1, mid;
    while (l<r){
        mid = (l+r)>>1;
        if (Check(mid)) l = mid+1;
        else r = mid;
    }
    if (l<m) printf("-1
%d", l);
    else printf("0");
}