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Return to Printing A.
Imagine you have a special keyboard with the following keys:
A
Ctrl+A
Ctrl+C
Ctrl+V
where CTRL+A, CTRL+C, CTRL+V each acts as one function key for “Select All”, “Copy”, and “Paste” operations respectively.
If you can only press the keyboard for N times (with the above four keys), please write a program to produce maximum numbers of A. If possible, please also print out the sequence of keys. That is to say, the input parameter is N (No. of keys that you can press), the output is M (No. of As that you can produce).
<syntaxhighlight lang="c" name="printa">
int main() {
unsigned int N, i, j; unsigned long long M; unsigned long long sol[1000]; unsigned long long buf[3]; unsigned int ind[1000]; printf("Enter N: "); scanf("%lu", &N); unsigned long long arr[3]; memset(arr, 0, 3*sizeof(arr[0])); memset(sol, 0, 1000*sizeof(sol[0])); /* arr[0] = 4; buf[0] = 1; arr[1] = 6; buf[1] = 2; arr[2] = 6; buf[2] = 3; */ arr[0] = 4; buf[0] = 1; arr[1] = 6; buf[1] = 2; arr[2] = 6; buf[2] = 3; unsigned index; char * out[] = { "A\n", "CTRL+A\n", "CTRL+C\n", "CTRL+V\n"}; unsigned int outf[1000]; unsigned p = 2, pp = 3; memset(outf, 0, 1000*sizeof(outf[0])); for(i=1; i<=N; i++) { if(i <= 6) { outf[i] = 0; sol[i] = i; ind[i] = 0; ind[6] = 3; } else { printf("\n\ni = %u\n", i); arr[0] = arr[1] + buf[1]; arr[1] = arr[2] + buf[2]; arr[2] = 2 * sol[i-3]; buf[0] = buf[1]; buf[1] = buf[2]; buf[2] = sol[i-3]; if(arr[0] > arr[1]) { if(arr[0] > arr[2]) { index = i-5; sol[i] = arr[0]; // outf[i] = outf[i-1]; //outf[i-1] = outf[i-2]; if(p > 3){ outf[p+1] = 1; outf[p+2] = 2; } // for(j = 0; j < buf[0]; j++) // outf[j] = 0; for(j = p+3; j <=i; j++) outf[j] = 3; ind[i] = p; if(ind[p] >= 3){ outf[ind[p]+1] = 1; outf[ind[p]+2] = 2; } for(j = ind[p]+3; j <= p; j++) outf[j] = 3; } else { index = i-3; sol[i] = arr[2]; outf[i] = 3; outf[i-1] = 2; outf[i-2] = 1; outf[i-1] = 2; outf[i-2] = 1; ind[i] = i-3; outf[ind[i-3]+1] = 1; outf[ind[i-3]+2] = 2; for(j = ind[i-3]+3; j < i-3; j++) { outf[j] = 3; } // p = pp; // pp = i -3; // printf("CTRL+A\nCTRL+C\nCTRL+V\n"); } } else if (arr[1] > arr[2]) { index = i-4; sol[i] = arr[1]; outf[pp+1] = 1; outf[pp+2] = 2; // for(j = 0; j < buf[1]; j++) // outf[j] = 0; for(j = pp+3; j <=i; j++) outf[j] = 3; ind[i] = pp; if(ind[pp] >= 3){ outf[ind[pp]+1] = 1; outf[ind[pp]+2] = 2; } for(j = ind[pp]+3; j <= pp; j++) outf[j] = 3; //outf[i-1] = outf[i-2]; //outf[i-2]= outf[i-3]; //printf("CTRL+V\n"); } else { index = i-3; sol[i] = arr[2]; outf[i] = 3; outf[i-1] = 2; outf[i-2] = 1; ind[i] = i-3; outf[ind[i-3]+1] = 1; outf[ind[i-3]+2] = 2; for(j = ind[i-3]+3; j < i-3; j++) { outf[j] = 3; } //p = pp; // pp = i -3; //printf("CTRL+A\nCTRL+C\nCTRL+V\n"); } // sol[i] = max(arr[0] , arr[1], arr[2]); printf("\n%lld %lld %lld\n", arr[0], arr[1], arr[2]); printf("%lld %lld %lld\n", buf[0], buf[1], buf[2]); printf("p = %u pp = %u\n", p, pp); for(j = 1; j <= i; j++) { // printf(out[outf[j]]); } p = pp; pp = i - 3; } } i = 1; while(i+1 < N && outf[i+1] != 2) { outf[i] = 0; i++; } printf("\n\nKey Sequences:\n\n"); // if(N <= 6) { // printf("A\nA\nA\nA\nA\nA\n"); } //else { for(i = 1; i <= N; i++) { printf(out[outf[i]]); } } for(i = 1; i <= N; i++) { printf("%u ",ind[i]); } printf("\nM = %llu\n", sol[N]);
}
</syntaxhighlight>