这里列出了一些难做的题或是易错的题，简单的我就没有过多的解释，但是如果大家有任何问题都可以私信我或是评论一下，我会尽量即时的解答问题或是疑问的。
int x=0;
for(int i=4*n; i>=1; i–) x=x+2*i;
O(n)
The loop runs O(n) times and does O(1) work per iteration.int z=0;int x=0;for (int i=1; i<=n; i=i*3){ z = z+5; z++; x = 2*x;
}
O(log n)
Think about the values of i as the loop progresses. It will take on the series of values 1, 3, 9, 27, 81, 243, …, 3k. Since i is tripling on each iteration, it takes on successive powers of three.
The loop clearly only does O(1) work per iteration, so the mainquestion here is how many total iterations there will be. The loopwill stop when i > n. If we let k be some arbitrary iteration of theloop, the value of i on iteration k will be 3k. The loop stops when3k > n, which happens when k > log3 n. Therefore, the number ofiterations is only O(log n)int y=0;for(int j=1; j*j<=n; j++)j <= (n)^1/2 y++;O(√n)Notice that j is still growing linearly, but the loop runs as long asj2 ≤ n. This means that as soon as j exceeds √ n, the loop willstop. Therefore, there will only be O(√n) iterations of the loop,and since each one does O(1) work, the total work done is O(√n)int b=0; //constantfor(int i=n; i>0; i–) for(int j=0; j<i; j++)
b=b+5;
O(n^2)The most accurate answer would be O(n2)int y=1;int j=0;for(j=1; j<=2n; j=j+2) y=y+i;int s=0;for(i=1; i<=j; i++)
s++;
O(n)int b=0;for(int i=0; i<n; i++) for(int j=0; j<i*n; j++) b=b+5;The inner loop will run 0 + n + 2n + 3n + 4n + … + n(n-1) = n(0 + 1+ 2 + … + n – 1) times, so the total work done is O(n3). Youshouldn’t multiply by the number of times the outer loop runs becauseyou’re already summing up across all iterations. The most accurateruntime would be O(n3)int t=0;for(int i=1; i<=n; i++) for(int j=0; j*j<4*n; j++) for(int k=1; k*k<=9*n; k++) t++;Look at the second loop. This actually runs O(√n) times using thesame logic as one of the earlier parts. That third inner loop alsoruns O(√n) times, and so the total work done will be O(n2)int a = 0;int k = n*n;while(k > 1){ for (int j=0; j<n*n; j++) { a++; } k = k/2;}The outer loop starts with k initialized to n2, but notice that k ishalved on each iteration. This means that the number of iterations ofthe outer loop will be log (n2) = 2 log n = O(log n), so the outerloop runs only O(log n) times. That inner loop does do O(n2) work, sothe total runtime is O(n^2 log n)int i=0, j=0, y=0, s=0;for(j=0; j<n+1; j++) y=y+j;for(i=1; i<=y; i++)
s++;
O(n^3)
，而只有在充满了艰辛的人生旅途中，始终调整好自己观风景的心态，
