HDU 4699
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Problem Description
Sample Input
8I 2I -1I 1Q 3LDRQ 2
Sample Output
23HintThe following diagram shows the status of sequence after each instruction:sum[x]数组存储前x个数的和
dp[x]数组存储前x个数中连续最大和 dp[x]=max(dp[x-1],sum[x]);
#include<stdio.h>#include<string.h>#define INF -10000000int s1[1000001],s2[1000001],dp[1000001],pos1,pos2;int sum[1000001];int Max(int a,int b){return a>b?a:b;}int main(){int cas;while(scanf("%d",&cas)!=EOF){pos1=0;pos2=0;dp[0]=-INF; //注意初始为 负很大sum[0]=0;while(cas--){char it[5];scanf("%s",it);if(it[0]=='I'){scanf("%d",&s1[++pos1]);sum[pos1]=sum[pos1-1]+s1[pos1];dp[pos1]=Max(dp[pos1-1],sum[pos1]);}if(it[0]=='D'){pos1--;}if(it[0]=='L'){if(pos1!=0)s2[++pos2]=s1[pos1--];}if(it[0]=='R'){if(pos2!=0){s1[++pos1]=s2[pos2--];sum[pos1]=sum[pos1-1]+s1[pos1];dp[pos1]=Max(dp[pos1-1],sum[pos1]);}}if(it[0]=='Q'){int x;scanf("%d",&x);printf("%d\n",dp[x]);}}} return 0;}
0 0
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