Codeforces Round #105 (Div. 2) ABCDE

A. Insomnia Cure

Ah can only speak English is too poor, one eye question I watched for three minutes.

Test instructions: Give 5 numbers k, l, m, n and D, 1~d can be k, l, m< /c10>, n the number of at least one divisible.

A: ...

Code:

#include <iostream>using namespacestd;intMain () {intA, B, C, D, N; CIN>> a >> b >> c >> D >>N; intAns =0;  for(inti =1; I <= N; ++i)if(i% a && i% b && I% c && I% d) + +ans; cout<< N-ans; return 0;}

B. Escape

Test instructions: or 5 numbers , v_{P, vd, T, F and C, Princess Speed is VP, the speed of the Dragon is VD, the princess first run time t, the Dragon began chasing the princess, chasing, the Princess threw a treasure, the dragon picked up back to the beginning and then took the time F to hide the treasure, ask at least how many treasures princess to run distance c.}

Puzzle: Analog bar, there is a trick bar, that is, if the VP>=VD direct output 0.

Code:

#include <iostream>using namespacestd;intMain () {intVP, VD, T, F, C; CIN>> VP >> VD >> T >> F >>C; DoubleA = VP *T; DoubleD = VD-VP; intAns =0; DoubleT1 = A/D; A+ = T1 *VP; if(VP >= VD) {cout <<'0';return 0; }  while(A <c) {ans++; A+ = VP * (f + A/VD); T1= A/D; A+ = T1 *VP; } cout<<ans; return 0;}

C. Terse Princess

Test instructions: Silly forced princess Choose husband, n candidates, each has a wealth value, when see a person than before all the people are rich, the princess shouted "Oh", when see a person's wealth than in front of all the wealth and more, the princess shouted "Wow" (at this time do not shout "Oh"), a total of n people, shout a "Oh" and B-Times "Wow" allows you to construct a wealth value sequence that is not built out on the output-1.

The puzzle: Probably think about to know, first a 1, then B "Wow", then a "Oh", and then all 1. There is a trick, if b=0, then the constructed array is 1 2 ... , 2 is going to be "Wow". (Div2 c I will not do, weak not ...)

Code (write a diff):

#include <iostream>#include<cstdio>using namespacestd;intMain () {intA, b, N; CIN>> N >> a >>b; inti =0; intans[ $]; ans[0] =1; if(b) { for(i =1; I <= b; ++i) {ans[i]= ans[i-1] *2; if(Ans[i] >50000) {printf ("-1"); return 0; }        }         for(; I <= b+a; + +)i) {ans[i]= ans[i-1] +1; }         for(; i < n; + +)i) {ans[i]=1; }    } Else {        if(A && a >= n1) {printf ("-1"); return 0; } ans[1] =1;  for(i =2; I <= A +1; ++i) {ans[i]= ans[i-1] +1; }         for(; i < n; ++i) Ans[i] =1; } printf ("%d", ans[0]);  for(inti =1; I < n; ++i) printf ("%d", Ans[i]); return 0;}

D. Bag Of Mice

Test instructions: Or a silly princess, Helong play games. There is a pocket, there are w white mice and b black mice, two people take turns to catch the mouse, the first to catch microscope win, the Dragon will scare off a mouse after a time, scare away the mouse is not caught. Princess first Catch, if no one caught microscope is dragon win. The probability of seeking a princess to win.

The more simple probability DP. Positive push from the initial state.

Code:

#include <iostream>#include<cstdio>using namespacestd;Const intN =1005;Doubledp[n][n][2];intMain () {intA, B; CIN>> a >>b; dp[a][b][1] =1;  for(inti = A; I >=0; --i) { for(intj = b; J >=0; --j) {if(j) dp[i][j-1][2] + = dp[i][j][1] * J/(i+j); if(J >1) dp[i][j-2][1] + = dp[i][j][2] * J/(I+J) * (J-1)/(i+j-1); if(i && J) dp[i-1][j-1][1] + = dp[i][j][2] * J/(I+J) * I/(i+j-1); }    }    DoubleAns =0;  for(inti = A; I >=0; --i) for(intj = b; J >=0; --j)if(i) ans + = dp[i][j][1] * I/(i+j); printf ("%.9f", ans); return 0;}

E. Porcelain

Test instructions: N shelves, there are some porcelain on each shelf, for each shelf can only take one from the side (left or right), ask M times to fetch the total value of porcelain.

Puzzle: Similar to 01 backpack. Preprocess each shelf with 1~c (assuming that the shelf has a total of C porcelain) The maximum value of a porcelain. Each shelf can only be selected once. Look at the code specifically.

Code:

#include <iostream>#include<cstdio>using namespacestd;Const intN = the;intA[n], c[n];intV[n][n];intdp[10005];intL[n], r[n];intMain () {intN, M; CIN>> N >>m;  for(inti =1; I <= N; ++i) {cin>>C[i];  for(intj =1; J <= C[i]; ++J) Cin >>A[j];  for(intj =1; J <= C[i]; ++j) {L[j]= l[j-1] +A[j]; R[J]= r[j-1] + a[c[i]-j+1]; }         for(intj =1; J <= C[i]; ++j) for(intK =0; K <= J; ++k) V[i][j]= Max (V[i][j], l[k] + r[j-K]); }     for(inti =1; I <= N; ++i) for(intj = m; J >=0; --j) for(intK =0; K <= C[i] && k <= J; ++k) Dp[j]= Max (Dp[j], dp[j-k] +V[i][k]); cout<<Dp[m]; return 0;}

Codeforces Round #105 (Div. 2) ABCDE