QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#584021#9346. Binary Numberskevinshan#WA 24ms48960kbC++142.0kb2024-09-23 04:07:242024-09-23 04:07:24

Judging History

你现在查看的是最新测评结果

  • [2024-09-23 04:07:24]
  • 评测
  • 测评结果:WA
  • 用时:24ms
  • 内存:48960kb
  • [2024-09-23 04:07:24]
  • 提交

answer

#include<bits/stdc++.h>
#define SZ(x) ((int)x.size())
#define FOR(i,a,b) for (int i=a;i<=b;++i)
#define FORD(i,a,b) for (int i=a;i>=b;--i)
using namespace std;

const int MO=100000007;
int h[2222222];
int N,m,L[2222222],R[2222222];
int f[140000][18][18];
int F(int x,int y){
    return m-1-h[x^y];
}
void doit(){
    scanf("%d%d",&m,&N);
    FOR(i,1,N){
        scanf("%d%d",&L[i],&R[i]);
        // FOR(j,L[i],R[i])
        //     bl[j]=i;
    }
    
    FOR(i,0,N)
        FOR(j,0,m)
            FOR(k,0,m)
                f[i][j][k]=0;
    f[0][m][0]=1;
    FOR(i,0,N-1)
        FOR(j1,0,m)
            FOR(j2,0,m){
            //f[i][j]
            if (f[i][j1][j2]==0) continue;
            if (i==1 && j1==2 && j2==0){
                int tt=1;
            }
            FOR(k,L[i+1],R[i+1])
                if (i==0 || (F(k,R[i]) <= j1 && F(k,L[i+1]) >= j2)){
                    int p1 = m-1-h[k^R[i+1]];
                    int p2;
                    if (i<N-1) p2 = m-1-h[k^(1+R[i+1])];
                    else p2 = 0;
                    (f[i+1][p1][p2]+=1ll*f[i][j1][j2]*k%MO)%=MO;
            //     if ()
            //     //R[i] qian j
            //     int t = h[R[i]^k];
            //     bool ok =0;
            //     if (t<m-j)
            //         ok=1;
            //     else{
            //         int ll = ((R[i]>>t)<<t);
            //         int rr = (((R[i]>>t)+1)<<t)-1;
            //         if (max(ll,L[i+1])>min(rr,R[i+1]))
            //             ok=1;
            //     }
            //     if (ok)
            //         (f[i+1][m-1-h[k^R[i+1]]]+=f[i][j] * k)%=MO;
            }
        }
    long long ans=0;
    FOR(j1,0,m) FOR(j2,0,m)
    (ans+=f[N][j1][j2])%=MO;
    printf("%d",(int)ans);
}
int main(){
    m=17;
    h[0]=-1;
    FOR(i,1,(1<<m)-1)
        FORD(j,m-1,0)
            if ((i>>j)&1){
                h[i]=j;
                break;
            }
    int T;
    scanf("%d",&T);
    while (T--) doit();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 2ms
memory: 10028kb

input:

1
2 2
0 1
2 3

output:

5

result:

ok single line: '5'

Test #2:

score: -100
Wrong Answer
time: 24ms
memory: 48960kb

input:

20
4 6
0 1
2 3
4 6
7 7
8 11
12 15
9 39
0 31
32 47
48 51
52 63
64 87
88 92
93 95
96 127
128 143
144 159
160 167
168 175
176 191
192 207
208 255
256 263
264 271
272 283
284 287
288 289
290 295
296 303
304 319
320 351
352 357
358 367
368 375
376 383
384 391
392 399
400 403
404 407
408 415
416 443
444 4...

output:

4309202765175780424846877167144008233504028188720873858312833065791763968656974862319390622687359826711686720548893476832017902854705494

result:

wrong answer 1st lines differ - expected: '430920', found: '430920276517578042484687716714...6720548893476832017902854705494'