CF1151div2(Round 553)

思路题大赛

A

少考虑了一种情况,到死没想到

B

貌似我随机化50000次,没找到就无解貌似也过了

感觉随随便便乱搞+分类讨论都可以过的样子

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 505;
inline int read(){
    int v = 0,c = 1;char ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-') c = -1;
        ch = getchar();
    }
    while(isdigit(ch)){
        v = v * 10 + ch - 48;
        ch = getchar();
    }
    return v * c;
}
int a[N][N];
int b[N][N];
int tot;
int s[N];
int n,m;
inline bool pan(){
    for(int i = 1;i <= n;++i)
        for(int j = 1;j <= n;++j)
            if(b[i][j] != 1) return 0;
    return 1;
}
inline bool check(){
    if(pan()){
        if(n & 1){
            for(int i = 1;i <= n;++i) s[++tot] = 1;
            return 1;   
        }
        else return 0;
    }
    
}
int main(){
    srand(time(0));
    n = read(),m = read();
    for(int i = 1;i <= n;++i) for(int j = 1;j <= m;++j) a[i][j] = read();
    for(int t = 1;t <= 50000;++t){
        int now = 0;
        for(int i = 1;i < n;++i) s[i] = rand() % m + 1,now ^= a[i][s[i]];
        for(int i = 1;i <= m;++i){
            if((now ^ a[n][i]) != 0){
                s[n] = i;
                printf("TAK\n");
                for(int j = 1;j <= n;++j) printf("%d ",s[j]);
                return 0;
            }
        }
    }
    printf("NIE\n");
    return 0;
}

C

~~辣鸡CF连__int128都不支持~~

我们将所有的区间分成log块去考虑

每一块的内部其实都是等差数列

我们可以用等差数列求和公式

对于\(L,R\)就求一下前缀和

另外项数可能很大,一定要%mod（我就是因为这个WA的）

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const LL mod = 1e9 + 7;
inline LL read(){
    LL v = 0,c = 1;char ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-') c = -1;
        ch = getchar();
    }
    while(isdigit(ch)){
        v = v * 10 + ch - 48;
        ch = getchar();
    }
    return v * c;
}
LL L,R;
LL sum[129];
LL shou[129];
inline LL quick(LL x,LL y){
    LL res = 1;
    while(y){
        if(y & 1) res = res * x % mod;
        y >>= 1;
        x = x * x % mod;    
    }   
    return res;
}
LL inv2 = quick(2,mod - 2);
inline LL work(LL x){
    if(x == 0) return 0;
    LL cnt = 0;
    LL gg = 0;
    LL ans = 0;
    while(gg + (1ll << cnt) <= x){
        ans = (ans + sum[cnt]) % mod;
        gg += (1ll << cnt);
        cnt++;  
    }
    if(gg != x){
        LL rest = (x - gg) % mod;
        LL rail = (shou[cnt] + (rest - 1) * 2 % mod) % mod;
        ans = (ans + (shou[cnt] + rail) % mod * rest % mod * inv2 % mod) % mod;
    }
    return ans;
}
int main(){
    L = read(),R = read();
    LL now = 0;
    LL base = 0;
    LL ji = 1;
    LL ou = 2; 
    long long rr = R;
    do{
        rr -= (1ll << base);
    //  printf("%lld\n",rr);
        if(base & 1){
            shou[base] = ou;
            sum[base] = (ou + (ou + 2 * (quick(2,base) - 1) % mod) % mod) % mod * quick(2,base) % mod * inv2 % mod;
            ou = (ou + 2 * (quick(2,base)) % mod) % mod;
        }
        else{
            shou[base] = ji;
            sum[base] = (ji + (ji + 2 * (quick(2,base) - 1) % mod) % mod) % mod * quick(2,base) % mod * inv2 % mod;
            ji = (ji + 2 * (quick(2,base)) % mod) % mod;
        }
        base++;
    }while(rr > 0);
//  for(int i = 0;i <= base;++i) printf("%lld\n",(long long)(sum[i]));
    long long ans = ((work(R) - work(L - 1) + mod) % mod + mod) % mod;
    printf("%lld\n",(long long)ans); 
    return 0;
}

D

化式子可以发现,只和\(a_i-b_i\)的值有关

然后就快乐排序算贡献

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const LL mod = 1e9 + 7;
inline LL read(){
    LL v = 0,c = 1;char ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-') c = -1;
        ch = getchar();
    }
    while(isdigit(ch)){
        v = v * 10 + ch - 48;
        ch = getchar();
    }
    return v * c;
}
LL L,R;
LL sum[129];
LL shou[129];
inline LL quick(LL x,LL y){
    LL res = 1;
    while(y){
        if(y & 1) res = res * x % mod;
        y >>= 1;
        x = x * x % mod;    
    }   
    return res;
}
LL inv2 = quick(2,mod - 2);
inline LL work(LL x){
    if(x == 0) return 0;
    LL cnt = 0;
    LL gg = 0;
    LL ans = 0;
    while(gg + (1ll << cnt) <= x){
        ans = (ans + sum[cnt]) % mod;
        gg += (1ll << cnt);
        cnt++;  
    }
    if(gg != x){
        LL rest = (x - gg) % mod;
        LL rail = (shou[cnt] + (rest - 1) * 2 % mod) % mod;
        ans = (ans + (shou[cnt] + rail) % mod * rest % mod * inv2 % mod) % mod;
    }
    return ans;
}
int main(){
    L = read(),R = read();
    LL now = 0;
    LL base = 0;
    LL ji = 1;
    LL ou = 2; 
    long long rr = R;
    do{
        rr -= (1ll << base);
    //  printf("%lld\n",rr);
        if(base & 1){
            shou[base] = ou;
            sum[base] = (ou + (ou + 2 * (quick(2,base) - 1) % mod) % mod) % mod * quick(2,base) % mod * inv2 % mod;
            ou = (ou + 2 * (quick(2,base)) % mod) % mod;
        }
        else{
            shou[base] = ji;
            sum[base] = (ji + (ji + 2 * (quick(2,base) - 1) % mod) % mod) % mod * quick(2,base) % mod * inv2 % mod;
            ji = (ji + 2 * (quick(2,base)) % mod) % mod;
        }
        base++;
    }while(rr > 0);
//  for(int i = 0;i <= base;++i) printf("%lld\n",(long long)(sum[i]));
    long long ans = ((work(R) - work(L - 1) + mod) % mod + mod) % mod;
    printf("%lld\n",(long long)ans); 
    return 0;
}

E

首先一个小\(trick\)

连通块数 = 点数 - 边数

所以答案变成了点数之和减去边数之和

对于一个点\(i\),他的贡献应该是
\[ a_i*(n - a_i + 1) \]
就是左端点和右端点的取值都要合法

一条边存在仅当他链接的两个点都存在
\[ min(a_i,a_{i + 1}) * (n - max(a_i,a_{i + 1}) + 1) \]

原文地址：https://www.cnblogs.com/wyxdrqc/p/11402982.html