3 100 10000 100 101 102
35
1 34
0
3
3
3
...
3
99
100

i guess the answer is 201(1-35-34)
can DP give this answer? i think only ShortestPath can give this answer.

WA#8
#include<iostream>;
using namespace std;

int main()
{
    int a[100002],i,n,record,maxrecord,irecord,imax,nmax;
    cin >> n;
    cin >>a[1];
    imax=1;
    nmax=a[1];
    record=1;
    irecord=1;
    maxrecord=1;
    for(i=2;i<=n;i++)
    {
        cin >> a[i];
        if (a[i]>nmax)
        {
            ++record;
            if (maxrecord<record)
            {
                maxrecord=record;
                irecord=imax;
            }
        }
        else
        {
            imax=i;
            nmax=a[i];
            record=1;
        }
    }
    cout << irecord;
    system ("PAUSE");
    return 0;
}

Дайте, пожалуйста, подсказку...

#include<stdio.h>
#include<math.h>
#include<map>
using namespace std;
map<unsigned long long,int> M;

int Prime[30]={0,2,3,5,7},Pct=5;
void genprime(int);

int expP[30]={70},solVal[500],ctsol;
unsigned long long solKey[500];
void gensol(unsigned long long,int);
void arrange();

main()
{
    genprime(30);
    gensol(1,1);
    arrange();
    //for (int i=1;i<=ctsol;i++) printf("%lld %d\n",solKey[i],solVal[i]);

    int Test,id;
    unsigned long long n;
    scanf("%d",&Test);
    for (int t=1;t<=Test;t++)
    {
        scanf("%llu",&n);
        id=ctsol;
        while (solKey[id]>n) id--;
        printf("%llu %d\n",solKey[id],solVal[id]);
    }
}
void genprime(int K)
{
    int i,j;
    for (i=11;Pct<K;i+=2) {
        for (j=1;Prime[j]*Prime[j]<=i;j++)
            if (i%Prime[j]==0)
                break;
        if (Prime[j]*Prime[j]>i)
            Prime[Pct++]=i;
    }
    //for (int i=1;i<Pct;i++) printf("%d\t",Prime[i]);
}
void gensol(unsigned long long x,int pID)
{
    int i,pro=1;
    for (i=1;i<30;i++)
        pro*=expP[i]+1;
    M[x]=pro;

    if (log10(x)+log10(Prime[pID])<=18.3 && expP[pID]<expP[pID-1]) {
        expP[pID]++;
        gensol(x*Prime[pID],pID);
        expP[pID]--;
    }
    if (log10(x)+log10(Prime[pID+1])<=18.3 && expP[pID]!=0) {
        expP[pID+1]++;
        gensol(x*Prime[pID+1],pID+1);
        expP[pID+1]--;
    }
}
void arrange()
{
    map<unsigned long long,int>::iterator it;
    for (it=M.begin();it!=M.end();it++) {
        if (it->second > solVal[ctsol]) {
            ctsol++;
            solKey[ctsol]=it->first;
            solVal[ctsol]=it->second;
        }
    }
}

My source is consisted of 8 cycles from zero to nine and one counting and works for 0.356 or something like that.

I don't understand what the output means?

thanks for help!

For WA #6:
It is not necessarily the three consecutive numbers to be equal. So if you have the test:
3
1 2 3
the answer should be 6 2.

For WA #2:
The consecutive numbers, which you sum, have to be exactly three.

can any1 tell me what is solution for 4x4,4x5 and 4x6, also some hints would be nice :D
Thanks

I've no idea.Could anybody help me?

According to the problem statment 1 ≤ L1 < L2 < L3 ≤ 10 ** 9, 1 ≤ C1 < C2 < C3 ≤ 10 ** 9 but test #10 fails with integers (32-bit) for Li and Ci.

var n,k,x: integer;
prov,res:real;
begin
read(n,k);
  x:=n*2;
    res:=x/k;
      res:=trunc(res); writeln(res);
end.

just dfs and pre calculate gcd.

As far as I can see, there are not so much test cases published in the problem's forum at this moment. So I think I should fill the gap. Here are tests and my AC program's answers. I hope these will help.

6 3
1 2 3
answer: 9

9 3
1 2 3
answer: 15

9 3
2 3 4
answer: 15

9 1
1
answer: 0

15 6
1 2 3 4 5 6
answer: 81

15 6
2 3 4 5 6 7
answer: 63

15 7
1 2 3 4 5 6 7
answer: 91

999 500
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 5
7 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83
84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107
108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127
128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147
148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167
168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187
188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207
208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227
228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247
248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267
268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287
288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307
308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327
328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347
348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367
368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387
388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407
408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427
428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447
448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467
468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487
488 489 490 491 492 493 494 495 496 497 498 499 500
answer: 497503


900 400
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83
84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107
108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127
128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147
148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167
168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187
188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207
208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227
228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247
248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267
268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287
288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307
308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327
328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347
348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367
368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387
388 389 390 391 392 393 394 395 396 397 398 399 400 401
answer: 279400

It works on my laptop. I have Pascal ABC.
var A:array[1..1000]of string;
B:array[1..3]of integer;
c,d:integer;
begin
readln(c);
for d:=1 to c do begin
readln(A[d]);
if A[d]='Emperor Penguin'then B[1]+=1;
if A[d]='Little Penguin'then B[2]+=1;
if A[d]='Macaroni Penguin'then B[3]+=1;
end;
d:=max(max(B[1],B[2]),B[3]);
if d=B[1]then write('Emperor Penguin');
if d=B[2]then write('Little Penguin');
if d=B[3]then write('Macaroni Penguin');
end.

Edited by author 23.08.2013 19:43

Edited by author 23.08.2013 19:43

What's mean this problem?

 Everybody who knows quick sort will recognize that the tsar's program is quick sort of N numbers. Let we have a sorted array with n numbers. Let l-r+1=k. Then function P() will increase the value of c by k+1 because it will visit all x elements to distribute them into part smaller than or equal to x and part greater or equal to x. That increases c by k. But the i and j iterators must be equal so there is also one additional increase of c. Because we have a sorted array function P() will partition him into part consisting of 1 element and part consisting of l-r elements. The part consisting of 1 element will be ignored and there won't be any increase of c. In the part with l-r elements it will do the same - partitioned it into 1 and l-r-1. So let begin with l-r+1=N. Then the function Q() will go to l-r+1=N-1 and so on to l-r+1=2. For those calls of function Q() c will be increased with:
N+1 + N + N-1 + ... + 5 + 4 + 3 = (N+1)*(N+1+1)/2 - 1 - 2 = N*N + 3*N + 2/2 - 3 =
=N*N + 3*N + 2 - 3*2/2 = N*N + 3*N - 4 / 2.
So if we give a sorted array to the tsar's program it will print what we want. What is more, we prove that for sorted array quick sort is very bad algorithm with such bad complexity.

Note: If the median element was chosen in a different way the solution won't be so easy. But it can be done if we try all permutations of array with elements from [1;n] and check for c.

Edited by author 21.08.2013 22:19

please give me some tests