动态规划,而已! CodeForces 433B - Kuriyama Mirai's Stones
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
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She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n),
and you should tell her
.
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Let ui be the
cost of the i-th cheapest stone (the cost that will be on the i-th
place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n),
and you should tell her
.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
6 6 4 2 7 2 7 3 2 3 6 1 3 4 1 1 6
24 9 28
4 5 5 2 3 10 1 2 4 2 1 4 1 1 1 2 1 4 2 1 2 1 1 1 1 3 3 1 1 3 1 4 4 1 2 2
10 15 5 15 5 5 2 12 3 5
Please note that the answers to the questions may overflow 32-bit integer type.
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#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; #define maxn 100006 __int64 sum; __int64 pp[maxn]={0},p[maxn]={0},liu[maxn],xp[maxn]; int main() { int i,j,k; int t,n,m; int l,r; while(scanf("%d",&n)!=EOF) { p[0]=0; for(i=1;i<=n;i++) { scanf("%I64d",&liu[i]); xp[i]=liu[i]; p[i]=p[i-1]+liu[i]; } xp[0]=0; sort(xp,xp+n+1); for(i=1;i<=n;i++) pp[i]=pp[i-1]+xp[i]; scanf("%d",&t); while(t--) { scanf("%d",&m); if(m==1) { scanf("%d%d",&l,&r); sum=p[r]-p[l-1]; printf("%I64d\n",sum); } else if(m==2) { scanf("%d%d",&l,&r); sum=pp[r]-pp[l-1]; printf("%I64d\n",sum); } } } return 0; }看出bug就讲吧,谢谢;
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